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# Summing a Curious, Slowly Convergent Series

Authors:
• Abu Dhabi Investment Authority

## Abstract

If we delete from the harmonic series all terms whose denominators contain any pattern of digits such as "314159", the remaining terms form a convergent series. Ever since Kempner proved the first such surprising result in 1914, it has been difficult to compute the sums of these series. The convergence is so slow that even direct summation using all the computers in the world would be of no use. We present a method for efficiently computing these sums. For example, the sum of the series whose denominators omit "314159" is approximately 2302582.33386. We also explain why this sum is so close to 106 log(10).
Summing a Curious,
Slowly Convergent Series
Thomas Schmelzer and Robert Baillie
1. INTRODUCTION AND HISTORY. The harmonic series
s=1
1
s
diverges. Suppose we delete from the series all terms whose denominators, in base 10,
contain the digit 9. Kempner [13] proved in 1914 that the remaining series converges.
We also get convergent series by deleting terms whose denominators contain any digit
or string of digits, such as “42”, or “314159”.
However, these series converge so slowly that calculating their sums directly is out
of the question. Here, we describe an algorithm to compute these and related sums
to high precision. For example, the sum of the series whose denominators contain no
“314159” is approximately 2302582.33386. We explain why this sum is so close to
10
6
log 10 by developing asymptotic estimates for sums that omit strings of length n,
as n approaches inﬁnity.
At ﬁrst glance, it seems counter-intuitive that merely omitting the terms 1/9, 1/19,
1/29,... from the harmonic series would produce a convergent series. It appears that
we are removing only every tenth term from the harmonic series. If that were the case,
then the sum of the remaining terms would indeed diverge.
This series converges because in the long run, we in fact delete almost everything
from the harmonic series. We begin by deleting 1/9, 1/19, 1/29,... . But when we
reach 1/89, we delete 11 terms in a row: 1/89, then 1/90 through 1/99. When we
reach 1/889, we delete 111 terms in a row: 1/889, then 1/890 through 1/899, and
ﬁnally 1/900 through 1/999.
Moreover, the vast majority of integers of, say, 100 digits contain at least one “9”
somewhere within them. Therefore, when we apply our thinning process to 100-digit
denominators, we will delete most terms. Only 8 × 9
99
/(9 × 10
99
) 0.003% of terms
with 100-digit denominators will survive our thinning process. Schumer [14] argues
that the problem is that we tend to live among the set of puny integers and generally
ignore the vast inﬁnitude of larger ones. How trite and limiting our view!
We can paraphrase Kempner’s argument as follows. There are 8 × 9
i1
integers
with i digits that do not contain a “9”. Their reciprocals are all at most 1/10
i1
,sothe
sum of their reciprocals is at most 8 × (9/10)
i1
. Summing these numbers over i gives
a convergent geometric series that converges to 80. This is an upper bound of the sum
of the reciprocals of integers not containing a “9”.
Kempner’s reasoning and convergence result (but not his upper bound) apply to any
digit in any base. That is, if d is a digit in base B, then if we delete from the harmonic
series all terms that contain the base-B digit d, we likewise get a convergent series.
We can use this fact to show that deleting terms that contain any ﬁxed string of digits
also gives a convergent series.
Also, there is a connection between the set of numbers that contain the decimal
string “42” and the set of numbers that, in base 100, have a digit equal to 42. The
June–July 2008]
SUMMING A SLOWLY CONVERGENT SERIES 525
second is a proper subset of the ﬁrst. For example, (decimal) 1942 and 4219 have a
base-100 digit equal to 42, but 1429 does not.
Theorem 1. Let X be a string of n base-10 digits. Then if we delete from the harmonic
series all terms that contain X , the resulting series converges.
Proof. We may interpret the n-digit string X as a single digit in base 10
n
.LetY be
the set of all numbers in base 10
n
that contain the digit X . Deleting the terms whose
denominators are in
Y gives a convergent series. But if we delete the terms whose
decimal value contains X, we are deleting all elements of
terms. It therefore follows that if we delete terms from the harmonic series that contain
the base-10 digit string X, the remaining series also converges.
Once a series is known to converge, the natural question is, “What is its sum?”
Unfortunately, these series converge far too slowly to compute their sums directly
[12, 13].
The problem has attracted wide interest through the years in books such as [1,
p. 384], [4, pp. 81–83], [9, pp. 120–121] and [10, pp. 31–34]. The computation of
thesesumsisdiscussedin[2], [7], and [17]. Fischer computed 100 decimals of the
sum with “9” missing from the denominators, but his method does not readily gener-
alize to other digits. His remarkable result is that the sum is
β
0
ln 10
n=2
10
n
β
n1
ζ(n)
where β
0
= 10 and the remaining β
n
values are given recursively by
n
k=1
n
k
(10
nk+1
10
k
+ 1
nk
= 10(11
n
10
n
).
Trott [16, pp. 1281–1282] has implemented Fischer’s algorithm using Mathematica.
In 1979, Baillie [2] published a method for computing the ten sums that arise when
we delete terms containing each of the digits “0” through “9”. The sum with “9”
deleted is about 22.92067. But the sum of all terms with denominators up to 10
29
still differs from the ﬁnal sum by more than 1.
In order to compute sums whose denominators omit strings of two or more digits,
we must generalize the algorithm of [2]. We do that here. We will show how to compute
sums of 1/s where s contains no odd digits, no even digits, or no strings like “42” or
“314159” or even combinations of those constraints.
Recently the problem has attracted some interest again. The computation of the
sum of 1/s where s does not contain “42” is a problem suggested by Bornemann et
al. [5, p. 281]. Subsequently the problems related to those sums have been discussed
in a German-speaking online mathematics forum.
1
In 2005 Bornemann presented his
solution for the 42”-problem to Trefethen’s problem solving squad at Oxford. His
idea is very similar to the original approach of Baillie and is covered by our analysis.
Bold print indicates vectors, matrices, or tensors. Sets are in calligraphic print.
1
At http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=9875 a few partic-
ipants discuss the computation of the sum of 1/n where n does not contain an even digit. It seems their solution
combines a direct summation and Richardson extrapolation and is of limited accuracy.
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
2. RECURRENCE MATRICES. Let X be a string of n 1 digits. Let S be the set
of positive integers that do not contain X in base 10. We denote the sum by ,thatis,
=
sS
1
s
. (1)
If X is the single digit m, Baillie’s method partitions
S into subsets S
i
.Theith subset
consists of those elements of
S that have exactly i digits. The following recurrence
connects
S
i
to S
i+1
:
S
i+1
=
sS
i
{10s, 10s + 1, 10s + 2,... ,10s + 9}\{10s + m}
From this, a recurrence formula is derived that allows us to compute
sS
i+1
1/s
k
from the sums
sS
i
1/s
k
.Ifn > 1, there is no simple recurrence relation between S
i
and S
i+1
. However, we can further partition S
i
into subsets S
j
i
for j = 1, 2,... ,n in
a way that yields a recurrence between
S
j
i+1
and the sets S
1
i
,... ,S
n
i
.Oncewehave
done this, we have
=
n
j =1
i=1
sS
j
i
1
s
.
We let
S
j
i
be the set of all members of S
i
whose last j 1 digits match the ﬁrst j 1
digits of X, but whose last j digits do not match the ﬁrst j digits of X. Notice that if
j < n and d is any digit, then there is a number k such that appending the digit d to
an element of
S
j
i
leads to any element of S
k
i+1
. The same is true for j = n, except that
if d is the last digit of X then appending d to an element of
S
n
i
gives a number that
contains the string X , and is therefore not an element of
S. It is convenient to let S
j
be the union of S
j
i
, over all i . We will represent the partition and the corresponding
recurrence with an n × 10 matrix T.The( j, d) entry of T tells us which set we end
up in when we append the digit d to each element of
S
j
. In other words, if appending
the digit d to an element of
S
j
leads to an element of S
k
,thenT ( j, d) = k. If the digit
d cannot be appended because the resulting number would not be in
S,thenweset
T( j, d) to 0.
Here is an example that shows how to compute the matrix T for a given string. Let
S be the set of integers containing no “314”. We partition S into three subsets: S
1
consists of the elements of S not ending in 3 or 31. S
2
is the set of elements of S
ending in 3. S
3
is the set of elements of S ending in 31. The following matrix T shows
what happens when we append the digits 0 through 9 to
S
1
, S
2
,andS
3
.
T =
0123456789
1 1112111111
2
1312111111
3
1112011111
.
When we append a 3 to an element of
S
1
, we get an element of S
2
,sowesetT(1, 3) =
2. Appending any other digit to an element of
S
1
yields another element of S
1
,sofor
all other d,T(1, d) = 1. Consider elements of
S
2
. Appending a 1 yields an element of
S
3
; appending a 3 yields another element of S
2
. Appending any other digit yields an
June–July 2008]
SUMMING A SLOWLY CONVERGENT SERIES 527
element of S
1
. The only special feature of S
3
is that if we append a 4, we get a number
ending in 314, which is not in
S,sowesetT(3, 4) = 0.
Let us emphasize that the matrix T does not have to be induced by a string X. Indeed
our approach is more general. We can also solve a puzzle stated by Boas [3]asking
2
foranestimateofthesumof1/s where s has no even digits. Here it is enough to work
with one set
S but to forbid that an even integer can be attached. Hence T is
T =
0123456789
1 0101010101
.
Many more interesting examples are discussed in Section 5.
The recursive relations between the sets
S
1
,... ,S
n
may also be illustrated by
means of a directed graph. There is a directed edge from
S
i
to S
j
if by appending
an integer d to elements of
S
i
we end up in S
j
; see Figure 1. For further analysis
we assume only that the associated directed graph is strongly connected, that is, there
are directed paths from
S
i
to S
j
and S
j
to S
i
for any pair i = j. Graphs that are not
strongly connected can be induced by more exotic constraints but are not discussed
here.
S
3
S
1
S
2
S
3
S
1
S
2
Figure 1. Directed, strongly connected graphs. Left: Graph for the partition induced by the string “314”. Any
other string with three distinct digits would have the same graph. Right: Graph of the partition induced by the
string “333”. Note that the strings “332” and “323” would induce two alternative graphs not shown here. The
graph related to “233” would match the left graph.
In the next section, we show how T is used to compute
sS
i+1
1/s
k
from the values
of
sS
i
1/s
k
.
3. A RECURRENCE FORMULA. It may seem a bit odd to introduce sums of s
k
although only the case k = 1 is desired, but they enable us to exploit the recurrence
relations between the aforementioned sets
S
1
,... ,S
n
. The idea has been successfully
appliedin[2]. Let
j
i,k
=
sS
j
i
1
s
k
. (2)
Therefore the sum of s
k
where s ranges over all i-digit integers is an upper bound for
j
i,k
.Thereare10
i
10
i1
of these integers, each of which is at least 10
i1
. Therefore,
j
i,k
10
i
10
i1
1
10
(i1)k
=
9
10
(i1)(k1)
. (3)
2
Actually Boas is pointing to problem 3555 published in School Science and Mathematics, April 1975.
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
Using the new notation the problem is to compute
=
n
j =1
i=1
j
i,1
.
The recursive nature of the sets
S
j
i
is now used to derive recurrence relations for the
sums
j
i,k
. We introduce a tensor f of dimensions n × n × 10 with
f
jlm
=
1ifT(l, m) = j,
0otherwise.
(4)
This tells us in the sum below to either include a term ( f
jlm
= 1) or not include it
( f
jlm
= 0). Then
j
i,k
=
9
m=0
n
l=1
f
jlm
sS
l
i1
(
10s + m
)
k
. (5)
By construction of f the sum runs over all index pairs (l, m) such that T(l, m) = j,
which indicates that
10s + m | s
S
l
i1
S
j
i
.
Although equation (5) is the crucial link to the recurrence matrices it is of no compu-
tational use. It is still a direct computation of
j
i,k
. We should avoid summing over a
range of integers in subsets
S
l
i1
. Using negative binomial series [18] we observe
(10s + m)
k
= (10s)
k
w=0
(1)
w
k + w 1
w
m
10s
w
where 0
0
= 1 and deﬁne:
c(k,w)= (1)
w
k + w 1
w
.
We replace the term
(
10s + m
)
k
in equation (5) and get
j
i,k
=
9
m=0
n
l=1
f
jlm
sS
l
i1
(10s)
k
w=0
c(k,w)
m
10s
w
=
9
m=0
n
l=1
f
jlm
w=0
10
kw
c(k,w)m
w
sS
l
i1
s
kw
by reordering the sum. To simplify the notation we introduce
a(k,w,m) = 10
kw
c(k,w)m
w
June–July 2008] SUMMING A SLOWLY CONVERGENT SERIES 529
and write therefore
j
i,k
=
9
m=0
n
l=1
f
jlm
w=0
a(k,w,m)
l
i1,k+w
. (6)
Again it may seem odd that we have replaced the ﬁnite summation (5) in s by an
inﬁnite sum in w. But the inﬁnite sum decays so fast in w that truncation enables us to
approximate (6) much faster than evaluating the sums of equation (5).
4. TRUNCATION AND EXTRAPOLATION. We step into the numerical analysis
of the problem. So far we have only reformulated the summation by introducing the
partial sums
j
i,k
. The ultimate goal is the efﬁcient computation of =
i, j
j
i,1
.We
use the following scheme:
For i 3thesums
j
i,k
(k > 1 is needed in the next step) are computed directly as
suggested in equation (2).
For i > 3 a recursive evaluation of (6) is used. The indices i and w both run from 4
(resp. 0) to inﬁnity. For w we use a simple truncation by neglecting all terms
l
i1,k+w
smaller than an a priori given bound ε, or to be precise: we neglect all terms
l
i1,k+w
where the estimate (3) is smaller than ε,thatis,wherek + w is sufﬁciently large.
We do not bound the resulting error as a function of ε in order to give a rigorous
proof. Ignoring individual terms of a series once they are small is delicate. After all,
the subject of this work is the harmonic series, which is everyone’s favorite example
of a series that diverges even though its terms approach 0. However, for a convergent
series the truncated sum ignoring terms smaller than ε will converge towards the limit
for ε 0. The limit may be estimated by using extrapolation, after computing the
truncated sum for ε, ε
2
,andε
3
or even higher order.
For the same reason we can neglect for large i all contributions from terms
l
i1,k+w
with k + w>1. Once the algorithm has achieved that stage it is possible to apply
extrapolation. Equation (6) with w = 0andk = 1 reads in matrix form
1
i,1
.
.
.
n
i,1
9
m=0
10
1
f
11m
··· f
1nm
.
.
.
.
.
.
f
n1m
··· f
nnm

A
n
1
i1,1
.
.
.
n
i1,1
(7)
since a(1, 0, m) = 10
1
for all m.
The nonnegative matrix A
n
is a contraction, that is, all its eigenvalues lie within the
unit disc. The proof of this property reveals a link to graph theory.
Deﬁnition 1. The associated digraph of an n × n matrix A
n
is the directed graph with
vertices 1,... ,n and an edge from i to j if and only if A(i, j) = 0.
The associated digraph of A
T
n
with vertices 1,... ,n representing the sets S
j
,
j = 1,... ,n is exactly the graph illustrating the recurrence relations between the sets
S
j
as introduced above, see Figure 1, as A
T
(i, j) = A( j, i) = 0ifT(i, m) = j for
some m.
We have assumed that this graph is strongly connected and hence A
T
n
and A
n
are
irreducible
3
and therefore the Perron-Frobenius Theorem [11, Theorem 8.4.4] applies:
3
A matrix is reducible if and only if its associated digraph is not strongly connected [8,p.163ff.].
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
Theorem 2 (Perron-Frobenius). Let A be a nonnegative
4
and irreducible matrix.
Then
there is an eigenvalue λ
d
that is real and positive, with positive left and right
eigenvectors,
any other eigenvalue λ satisﬁes
|
λ
|
d
,
the eigenvalue λ
d
is simple.
The eigenvalue λ
d
is called the dominant eigenvalue of A.
It remains to show that the dominant eigenvalue λ
d
of A
n
is smaller than 1. Consider
the lth column of the matrix
f
11m
··· f
1nm
.
.
.
.
.
.
f
n1m
··· f
nnm
.
By deﬁnition of f (4) there is a 1 in row T(l, m) if T(l, m)>0. All other entries in the
column are zero. Therefore
a
l
1
1, where a
1
,... ,a
n
denote the columns of A
n
.
The existence of columns which have no nonzero entry implies that there are
columns of A
n
with
a
i
1
< 1. Let x be the right eigenvector of A
n
corresponding to
λ
d
. Applying the triangle inequality we conclude that
λ
d
x
1
=
A
n
x
1
n
i=1
|
x
i
|
a
i
1
<
n
i=1
|
x
i
|
=
x
1
(8)
and therefore λ
d
< 1. Note that we have used the fact that the elements x
i
of the
positive vector x cannot vanish for any index i. If all columns of the nonnegative
matrix A
n
satisfy
a
i
1
= 1, the matrix A
n
is called stochastic and (1,... ,1) is a left
eigenvector with eigenvalue 1. In this case A
n
is no longer a contraction. This situation
occurs if T contains no zero and hence no integers are deleted at all. This may serve
as an unusual explanation for the divergence of the harmonic series.
Having shown that the spectrum of A
n
lies within the unit disk, for large K we can
simplify
i=1
1
i+K ,1
.
.
.
n
i+K ,1
i=1
A
i
n
1
K ,1
.
.
.
n
K ,1
(9)
by a Neumann series
k=1
A
k
n
= (I A
n
)
1
I =: B
n
(10)
where I is an identity matrix of appropriate dimension. Hence
n
j =1
K
i=1
j
i,1
+
B
n
1
K ,1
.
.
.
n
K ,1
1
. (11)
4
A matrix is nonnegative if and only if every entry is nonnegative.
June–July 2008] SUMMING A SLOWLY CONVERGENT SERIES 531
Using the same idea we can also estimate the result of truncating the series after, say,
M + K digits using
M
k=1
A
k
n
= (I A
M+1
n
)(I A
n
)
1
I =: B
M
n
. (12)
5. EXAMPLES. We compute a few sums to a precision of 100 decimals, although if
desired, more could easily be obtained. All of these sums have been computed to even
higher precision, and in every case, the ﬁrst 100 decimals agree.
First, let us compute the sum originally considered by Kempner [13], namely, where
the digit “9” is missing from the denominators. Here, there is only one set
S = S
1
,
namely, the set of integers that do not contain a “9”. When we append a “9” to an
element of
S, we get a number not in S,soT(1, 9) = 0. When we append any other
digit, we get another element of
S, so for all other d, T(1, d) = 1:
T =
0123456789
1 1111111110
.
To 100 decimals, the sum is
22.92067 66192 64150 34816 36570 94375 93191 49447 62436 99848
15685 41998 35657 21563 38189 91112 94456 26037 44820 18989 ... .
In Section 2 we mentioned the sum of 1/s where s has no even digits. The sum is
3.17176 54734 15904 95722 87097 08750 61165 67970 50708 39628
57241 64186 89843 71376 88585 61926 68852 31080 74715 60454 ... .
Similarly, the sum over denominators with no odd digits can be found using the matrix
T =
0123456789
1 1010101010
.
In this case the sum is
1.96260 84129 94616 98515 91542 64737 29435 67128 30665 51443
53546 71522 23586 65760 95274 32927 13468 24171 73826 12704 ... .
In Section 2, we gave the matrix T that corresponds to the sum of 1/s where s has
no string “314”. The sum is:
2299.82978 27675 18338 45358 63536 11974 36784 61556 88394 19837
51645 98202 17625 43309 41712 63285 37992 24266 07454 90945 ... .
We can also compute the sum of 1/s where s has no string “314159”. Then the sum is:
2302582.33386 37826 07892 02375 60364 84435 61276 86862 90972 08627
80786 90557 30669 81792 73645 44979 47969 47311 14619 12012 ... .
This sum demonstrates the power of the technique presented here. Using (12) we cal-
culate that the partial sum of all terms whose denominators have 100000 or fewer digits
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
is about 219121.34825 .... This is still only 1/10th as large as the ﬁnal sum, and illus-
trates the futility of direct summation. Notice that this sum is close to 10
6
log 10. But
there is nothing special about “314159”; we observe similar results for other strings of
Let
S be the set of integers containing no “42”. Computing in this case is the
challenge recently posed by Bornemann et al. [5]. In this case
T =
0123456789
1 1111211111
2
1101211111
and the sum is given by
228.44630 41592 30813 25414 80861 26250 58957 81629 27539 83036
11859 13460 00045 28607 68650 21430 70480 46117 41443 21741 ... .
The next example combines various constraints and illustrates the ﬂexibility of our
approach. Let
S be the set of integers containing no even digits, no “55”, and no
“13579”. Then we deﬁne the following partition of
S. S
2
is the subset of numbers
ending in 5 (but not ending in 135),
S
3
is the set of numbers ending in 1, S
4
is the set
of numbers ending in 13,
S
5
is the subset of numbers ending in 135 and the elements
of
S
6
end in 1357. All remaining elements of S are in S
1
. Following those rules the
matrix T is given by
T =
0123456789
1 0301020101
2
0301000101
3
0304020101
4
0301050101
5
0301000601
6
0301020100
.
Here the sum is
3.09084 91496 53806 46825 46563 73157 80175 63888 91119 39765
22149 64013 36906 53946 19395 87929 18235 63131 88124 97325 ... .
Our algorithm can be easily generalized for other bases than 10. The sum of 1/s
where s has no “0” in base 100 is
460.52520 26385 12471 14293 67535 66415 29497 12569 09908 47934
06016 95672 87250 06818 86421 46967 22875 07176 27582 54794 ... .
All sorts of other experiments are possible. Interested readers might experiment by re-
moving from the harmonic series their personal favorite numbers, such as their birth-
days (as a single string, or as a set of three strings), their phone numbers, etc.
6. ASYMPTOTIC BEHAVIOR. We now discuss the asymptotic behavior of the
sums that arise when we remove from the harmonic series terms whose denomina-
tors contain a string X
n
of length n digits.
Data for several random strings of n = 20 digits are given in Table 1. It is striking
that, for each random string X
n
, the “normalized” sum,
X
n
/10
n
, is very close to
log 10 = 2.30258 50929 94045 68401 79914 ... .
June–July 2008]
SUMMING A SLOWLY CONVERGENT SERIES 533
Table 1. Sums for several random 20-digit strings
n String X
n
X
n
/10
n
20 21794968139645396791 2.30258 50929 94045 68397 52162
20 31850115459210380210 2.30258 50929 94045 68399 08824
20 67914499976105176602 2.30258 50929 94045 68401 09579
20 98297963712691768117 2.30258 50929 94045 68401 77079
Table 2 shows what happens for strings that consist of repeated patterns of sub-
strings. If a string consists of shorter substrings repeated two or more times, we deﬁne
the period of the string to be the length of that shortest substring. So, “11111” has
period 1, while “535353” has period 2.
Table 2. Sums for strings of periods 1, 2, and 5
n String X
n
X
n
/10
n
5 00000 2.55840 22969
10 0000000000 2.55842 78808 48652
15 000000000000000 2.55842 78811 04492 64603
20 00000000000000000000 2.55842 78811 04495 20443 88506
20 11111111111111111111 2.55842 78811 04495 20435 05433
20 44444444444444444444 2.55842 78811 04495 20442 19551
20 99999999999999999999 2.55842 78811 04495 20443 88506
4 4242 2.32542 92748
10 4242424242 2.32584 35278 62555
16 4242424242424242 2.32584 35282 76813 40798 19419
20 42424242424242424242 2.32584 35282 76813 82219 89695
20 09090909090909090909 2.32584 35282 76813 82221 85405
5 12345 2.30250 59575
10 1234512345 2.30260 81180 53596
15 123451234512345 2.30260 81190 75226 21998
20 12345123451234512345 2.30260 81190 75236 43628 01912
Here, the normalized sums appear to approach different limits. The limits do
not depend on which digits comprise the strings, but instead depend on the pe-
riods. When all digits are identical (period 1), the limit of the normalized sum
seems to be (10/9) log 10 = 2.55842 ... . When the period is 2, the limit seems
to be (100/99) log 10 = 2.32584 ... . For period 5, it’s (100000/99999) log 10 =
2.30260 ... .
We emphasize that we observe these same limits for other strings of digits. In the
limit, the particular digits in a string do not matter. What matters is the structure of the
strings.
Conjecture 1. Let (X
n
)
npN
be a sequence of strings, where each string X
n
has n
digits and has period p, where n is a multiple of p. Let
X
n
be the sum of 1/swheres
does not contain the substring X
n
.Then
lim
n→∞
X
n
/10
n
=
10
p
10
p
1
log 10.
534
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
The lack of a periodic pattern may be interpreted as the limit as p →∞.
Conjecture 2. Let (X
n
)
nN
be a sequence of strings, where each string X
n
has n digits
with no periodic pattern. Let
X
n
be the sum of 1/s where s does not contain the
substring X
n
.Then
lim
n→∞
X
n
/10
n
= log 10.
Our numerical experiments strongly support these conjectures. Rather than attempt
rigorous proofs, we prove a special case ( p = 1) of Conjecture 1. The proof of the
following Theorem reveals the strong link between the spectrum of B
n
and the limits
we observe in Tables 1 and 2.
Theorem 3. Let (X
n
)
nN
be a sequence of strings, where each string X
n
has n digits
that are all the same. Let
X
n
be the sum of 1/s where s does not contain the substring
X
n
.Then
lim
n→∞
X
n
/10
n
=
10
9
log 10.
We sketch the proof by a series of Lemmata analyzing the spectrum of the extrap-
olation matrices A
n
givenin(7)andB
n
linked via (10). A string X
n
= d ...d gives
rise to an n × n matrix A
n
via (7) of the form
A
n
=
1
10
99··· ··· 9
10 0
01
.
.
.
.
.
.
.
.
.
10
(13)
assuming that
S
1
is the set of integers not ending in the integer d, numbers in S
2
end
in d but not dd, numbers in
S
3
end in dd but not ddd, and so on.
The spectrum of A
n
lies within the unit disc. We can say much more about the
spectrum in this case. We need the Theorem of Gershgorin [11, Theorem 6.1.1] to
make more precise statements:
Theorem 4 (Gershgorin). Let A be a n × n matrix, and let
R
i
(A) =
n
j =1, j =i
!
!
a
ij
!
!
, 1 i n
denote the deleted absolute row sums of A. Then all the eigenvalues of A are located
in the union of the n discs
{z C |
|
z a
ii
|
R
i
(A)}, 1 i n. (14)
Furthermore, if a union of k of these n discs forms a connected region that is disjoint
from all the remaining n k discs, then there are precisely k eigenvalues of A in this
region.
June–July 2008]
SUMMING A SLOWLY CONVERGENT SERIES 535
Lemma 1. The matrix A
n
as deﬁned in (13) is diagonalisable. The eigenvalues of the
n × nmatrixA
n
are the solutions of the equation
λ
n
(1 λ) = 9/10
n+1
(15)
distinct from 1/10. There are exactly n 1 eigenvalues of A
n
contained in a disc of ra-
dius 1/10 centered at the origin. If λ is an eigenvalue of A
n
then
n1
n2
,... ,1)
T
is an eigenvector.
Proof. Let x
n
= (x
1
,... ,x
n
)
T
be an eigenvector of 10A
n
corresponding to the eigen-
value λ of 10A
n
.Then
10A
n
x
1
.
.
.
x
n
=
9
n
i=1
x
i
x
1
.
.
.
x
n1
= λ
x
1
.
.
.
x
n1
x
n
. (16)
Hence x
n1
= λx
n
, which implies x
n2
= λ
2
x
n
. Using induction we observe
x
nk
= λ
k
x
n
.
In particular x
1
= λ
n1
x
n
. Both matrices 10A
n
and A
n
share the same set of eigenvec-
tors. Hence
n1
n2
,... ,1)
T
is an eigenvector of A
n
by choosing x
n
= 1. The
ﬁrst row of (16) reveals that the characteristic equation of 10A
n
is
9
n1
i=0
λ
i
= λ
n
.
Hence λ = 1 is not an eigenvalue of 10A
n
.Forλ = 1 we can write
9
n1
i=0
λ
i
= 9
1 λ
n
1 λ
= λ
n
or
λ
n+1
10λ
n
+ 9 = 0. (17)
The n + 1 roots of this polynomial are all distinct as the roots of the derivative
(n + 1
n
10nλ
n1
are not roots of (17). That implies that each eigenvalue of 10A
n
is of algebraic multi-
plicity 1 and hence the matrix 10A
n
is diagonalisable. Clearly λ is an eigenvalue of A
n
if and only if 10λ is an eigenvalue of 10A
n
. Hence the eigenvalues of A
n
are precisely
the solutions other than 1/10 of the equation
10
n+1
λ
n+1
10
n+1
λ
n
=−9
which yields (15). The roots of the polynomial in (17) are the eigenvalues of the (n +
1) × (n + 1) companion matrix
536
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
H
n+1
=
10 0 ··· 0 9
1
.
.
.
.
.
.
10
.
We write H
n+1
= D
n+1
+ B
n+1
where D
n+1
= diag(10, 0,... ,0) and set H
ε
n+1
=
D
n+1
+ εB
n+1
for ε ∈[0, 1].Forallε<1 the union of the Gershgorin circles of
radius ε centered at the origin is disjoint from the disc centered at 10. Because the
eigenvalues of H
ε
n+1
are continuous functions of ε,therearen eigenvalues of H
1
n+1
lo-
cated within the closed unit disc, including the simple eigenvalue at 1. Hence there are
n 1 eigenvalues of A
n
within the disc of radius 1/10. Note that a direct application
of the Theorem of Gershgorin to A
n
does not yield the desired result.
Lemma 2. Let λ
n
be the dominant eigenvalue of A
n
deﬁned in (13).Then
lim
n→∞
λ
n
n
= 1.
Proof. Since λ
n
is an eigenvalue of A
n
, it is a solution of (15). The graph of λ
n
(1 λ)
is monotonically decreasing for λ>n/(n + 1).Butforλ = (10
n
1)/10
n
, λ
n
(1 λ)
is still larger than 9/10
n+1
, and hence 1
n
>(10
n
1)/10
n
.Takingthenth power
of all terms yields the desired result.
Lemma 3. Let
n
be the dominant eigenvalue of 1/10
n
B
n
deﬁned in (10) using A
n
given by (13).Then
lim
n→∞
n
=
10
9
. (18)
Proof. The dominant eigenvalue of B
n
is
(1 λ
n
)
1
1
where λ
n
is the dominant eigenvalue of A
n
. Hence
n
=
(1 λ
n
)
1
1
10
n
.
But (1 λ
n
) =
9
10
n+1
λ
n
n
as given by identity (15). Therefore
n
=
10
n+1
λ
n
n
9 × 10
n
1
10
n
Applying Lemma 2 ﬁnishes the proof.
All other eigenvalues of 1/10
n
B
n
are contained in a disc of radius
1
9×10
n
corre-
sponding to the eigenvalues of A
n
contained in a disc of radius 1/10.
The dominant eigenvector of 1/10
n
B
n
is the dominant eigenvector of 10A
n
.The
dominant eigenvalue of 10A
n
is approaching 10 and hence the dominant eigenvector
of 10A
n
is, using (16), converging towards (1, 1/10, 1/100,... ,1/10
n1
)
T
where we
have normalized this vector such that the ﬁrst component is 1.
June–July 2008]
SUMMING A SLOWLY CONVERGENT SERIES 537
Although the matrix 1/10
n
B
n
is invertible for any n we may regard the limit as an
operator of rank 1. The matrix-vector product 1/10
n
B
n
v
n
is in the limit a projection
of v
n
to (1, 1/10, 1/100,...1/10
n1
)
T
followed by a multiplication with 10/9. If a
vector v
n
is already aligned with the eigenvector the application of 1/10
n
B
n
v
n
boils
down to a multiplication by 10/9.
Lemma 4. Let B
n
as deﬁned in (10) using A
n
given by (13).Then
1
10
n
B
n
1
n1,1
.
.
.
n
n1,1
1
−−
n→∞
10
9
log 10
where
1
n1,1
,... ,
n
n1,1
are the partial sums of reciprocals of n 1 digit numbers
associated with the string X
n
= d ...d of length n.
Proof. The norm of the vectors
1
n1,1
.
.
.
n
n1,1
is exactly the sum of 1/s over all positive integers s with exactly n 1 digits. Not a
single integer has been deleted at this stage yet. Hence
1
n1,1
.
.
.
n
n1,1
1
−−
n→∞
"
10
n
10
n1
1
t
dt = log 10.
Next we claim that
lim
n→∞
1
1
n1,1
1
n1,1
.
.
.
n
n1,1
1
1/10
.
.
.
1/10
n1
1
= 0.
It is without doubt that the ﬁrst entry of the difference is indeed 0. The second entry
represents the ratio of the sum of the reciprocals of the n 1 digit numbers ending in
d but not dd and the sum of the reciprocals of the n 1 digit numbers not ending in
d. The ratio of the numbers of terms in the two sums is exactly 1/10. The ratio of the
sums approaches 1/10 since the numbers ending in d are equally distributed amongst
the numbers not ending in d. The same argument can be applied to any other row,
too. The coarse relative approximation of
n
n1,1
/
1
n1,1
by 1/10
n1
is irrelevant as
the maximal entry of the last n/2 rows of both vectors is converging towards 0. Hence
the vector
1
n1,1
.
.
.
n
n1,1
538
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
is turning into the direction of the dominant eigenvector of B
n
while approaching the
norm log 10.
Lemma 4 gives a new perspective on equation (11). There we ﬁx the truncation
parameter K to n 1 and introduce an error term e
n
compensating this step. Hence
X
n
10
n
=
1
10
n
n
j =1
n1
i=1
j
i,1
+
B
n
1
n1,1
.
.
.
n
n1,1
1
+ e
n
.
The largest terms collected in the error term are
j
n1,2
for j = 1,... ,n.Theirsum,
that is
n
j =1
j
n1,2
, is smaller than 9/10
n2
as shown by equation (3). Hence the terms
we neglect get exponentially smaller for increasing n. In the language of Section 4 we
are neglecting terms smaller than or equal to ε
n
=
1
n1,2
< 9/10
n2
.
We observe that for increasing n the double sum multiplied by 10
n
converges to
zero due to the slow growth of the sum. The error term scaled by 10
n
representing
the neglected sums meets the same fate. So Lemma 4 implies
lim
n→∞
X
n
10
n
= lim
n→∞
1
10
n
B
n
1
n1,1
.
.
.
n
n1,1
1
=
10
9
log 10.
This proof of Theorem 3 may serve as a blueprint for a proof of the two conjectures.
Such a proof following the footsteps of the Lemmata given above is ambitious as the
strings X
n
may give rise to a much larger class of matrices. The entries of the matrices
A
n
mildly depend on the order of the digits. We invite the readers of this MONTHLY
to prove Conjectures 1 and 2.
7. CONCLUSIONS. We have derived an algorithm that allows us to efﬁciently sum
the series that result when various constraints are used to delete terms from the har-
monic series. The algorithm uses truncation and extrapolation and avoids futile direct
summation. Embedding the problem in the language of linear algebra reveals an in-
teresting link to graph theory. The Perron-Frobenius Theorem and Gershgorin circles
enable us to present an asymptotic analysis for the special case when we delete integers
having n identical digits.
Our Mathematica implementation can be downloaded from the webpage of the ﬁrst
author.
5
The core of the program ﬁts on one page and produces 10 digits in less than
5 seconds. This is a model for good scientiﬁc computing that has recently been put
forward by Trefethen [15].
This paper provides many opportunities for further exploration. It might be good fun
to derive an algorithm for the inverse problem: what set of simple constraints might
one apply in order to make the sum as close as possible to a given number?
Are these sums rational, irrational, algebraic, or transcendental? All we know is that
a theorem of Borwein [6] implies that if the denominators that survive deletion consist
of a single nonzero digit d in base B, then the resulting sum is irrational. The N th
partial sum of the harmonic series is never an integer [10, p. 24]. Is that true for the
partial sums of the series we consider here?
5
http://web.comlab.ox.ac.uk/thomas.schmelzer
June–July 2008] SUMMING A SLOWLY CONVERGENT SERIES 539
ACKNOWLEDGMENTS. This note was inspired by Folkmar Bornemann who challenged Nick Trefethen
and his Problem Solving Squad at Oxford with his “42” problem. The ﬁrst author enjoyed discussions with
them and Gustav Holzegel, Nicolas Jeannequin, and Martin Stoll. We would like to thank the anonymous
referees for their helpful suggestions. T.S. acknowledges support by the Rhodes Trust.
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THOMAS SCHMELZER was a Rhodes Scholar at Balliol College, Oxford where he worked with
Lloyd N. Trefethen and ﬁnished a D.Phil. on matrix functions and parabolic PDEs in 2007. As an under-
graduate he studied mathematics and physics at Kaiserslautern and Zurich. He now works for Winton Capital
Management, a quantitative hedge fund. Whenever British weather permits he enjoys cycling and hiking. He
also tamed a hydra of singularities with Folkmar Bornemann in the October 2007 issue of this M
ONTHLY.
Winton Capital Management, Magdalen Centre, The Oxford Science Park, Oxford OX4 4GA, UK
thomas.schmelzer@gmail.com
ROBERT BAILLIE received B.S. and M.S. degrees in mathematics from the University of Illinois in Urbana
in 1970 and 1971. Since then, he has worked as a computer programmer. He is interested in number theory
and helped develop the Baillie-PSW primality test. He also enjoys doing experimental mathematics, using
computer software to shed light on new and unsolved problems.
Remcom, Inc., State College, Pennsylvania
rjbaillie@frii.com
540
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THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115
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... Given ǫ > 0, there exists a k-free Kempner set K such that H(K + 1) > H(S) − ǫ. One of the reasons to study Kempner sets is that machinery exists due to [SB08] to compute harmonic sums of Kempner sets with great precision. There is one small difficulty, in that Kempner sets include 0. Rather than exclude 0, we opt to increase our sets termwise by 1. ...
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Let $\mathscr{X}$ be the set of positive real sequences $x=(x_n)$ such that the series $\sum_n x_n$ is divergent. For each $x \in \mathscr{X}$, let $\mathcal{I}_x$ be the collection of all $A\subseteq \mathbf{N}$ such that the subseries $\sum_{n \in A}x_n$ is convergent. Moreover, let $\mathscr{A}$ be the set of sequences $x \in \mathscr{X}$ such that $\lim_n x_n=0$ and $\mathcal{I}_x\neq \mathcal{I}_y$ for all sequences $y=(y_n) \in \mathscr{X}$ with $\liminf_n y_{n+1}/y_n>0$. We show that $\mathscr{A}$ is comeager and that contains uncountably many sequences $x$ which generate pairwise nonisomorphic ideals $\mathcal{I}_x$. This answers, in particular, an open question recently posed by M. Filipczak and G. Horbaczewska.
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In diesem Beitrag geht es um die vierte Aufgabe der 2. Runde des Bundeswettbewerbs Mathematik 2006. Dabei soll der Wert der Summe von endlich vielen verschiedenen Stammbrüchen, deren Nenner jeweils mindestens eine der zehn möglichen Ziffern nicht enthalten, abgeschätzt werden. Je nach der gewählten Vorgehensweise gelingen dabei beachtliche Erkenntnisse über die auf diese Weise „ausgedünnte“ harmonische Reihe. Insbesondere werden zwei Möglichkeiten aufgezeigt, wie man die kleinstmögliche obere Schranke aller derartiger Summen beliebig genau bestimmen kann.
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Mathematica is today's most advanced technical computing system. It features a rich programming environment, two-and three-dimensional graphics capabilities and hundreds of sophisticated, powerful programming and mathematical functions using state-of-the-art algorithms. Combined with a user-friendly interface, and a complete mathematical typesetting system, Mathematica offers an intuitive easy-to-handle environment of great power and utility. "The Mathematica GuideBook for Symbolics" (code and text fully tailored for Mathematica 5.1) deals with Mathematica's symbolic mathematical capabilities. Structural and mathematical operations on single and systems of polynomials are fundamental to many symbolic calculations and they are covered in considerable detail. The solution of equations and differential equations, as well as the classical calculus operations (differentiation, integration, summation, series expansion, limits) are exhaustively treated. Generalized functions and their uses are discussed. In addition, this volume discusses and employs the classical orthogonal polynomials and special functions of mathematical physics. To demonstrate the symbolic mathematics power, a large variety of problems from mathematics and phyics are discussed. © 2006 Springer Science+Business Media, Inc. All rights reserved.