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Nilakantha's accelerated series for pi

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Abstract

We show how the idea behind a formula for π discovered by the Indian mathematician and astronomer Nilakantha (1445–1545) can be developed into a general series acceleration technique which, when applied to the Gregory-Leibniz series, gives the formula [...] with convergence as 13.5^{–n}, in much the same way as the Euler transformation gives [...] with convergence as 2^{–n}. Similar transformations lead to other accelerated series for π, including three "BBP-like" formulas, all of which are collected in an appendix. Optimal convergence is achieved using Chebyshev polynomials.
ACTA ARITHMETICA
* (201*)
Nilakantha’s accelerated series for π
by
David Brink (København)
1. Introduction. Unbeknownst to its European discoverers—Gregory
(1638–1675) and Leibniz (1646–1716)—the formula
(1) π
4=
X
n=0
(1)n
2n+ 1
had been found in India already in the fourteenth or fifteenth century. It
first appeared in Sanskrit verse in the book Tantrasangraha from about 1500
by the Indian mathematician, astronomer and universal genius Nilakantha
(1445–1545). Unlike Gregory and Leibniz, Nilakantha also gave approxima-
tions of the tail sums and found a more rapidly converging series,
(2) π
4=3
4+
X
n=0
(1)n
(2n+ 3)3(2n+ 3).
The reader is referred to Roy [14] for more details on this fascinating story.
We show here that (2) is the first step of a certain series transformation
that eventually leads to the accelerated series
(3) π=
X
n=0
(5n+ 3)n!(2n)!
2n1(3n+ 2)! ,
in much the same way as the difference operator leads to the Euler transform
(4) π=
X
n=0
2n+1n!n!
(2n+ 1)!.
We call (3) the Nilakantha transform of (1) and note that it converges
roughly as 13.5n, whereas the Euler transform converges as 2n. Applying
2010 Mathematics Subject Classification: Primary 65B10; Secondary 40A25.
Key words and phrases: series for π, convergence acceleration.
[1]
2 D. Brink
the Nilakantha transformation to the Newton–Euler formula [8]
(5) π
22=
X
n=0
(1)n(n1)/2
2n+ 1
gives the accelerated series (A.5) (see the Appendix), also with convergence
13.5n. Similar transformations of these and other formulas lead to other
accelerated series for πwhich are collected in the Appendix in a standardized
form, with (3) corresponding to (A.1).
Several of the formulas in the Appendix are well known from the litera-
ture. A series equivalent to (A.1) is attributed to Gosper in [2, eq. (16.81)].
The series (A.2) is due to Adamchik and Wagon [1] and is one of the simplest
of all “BBP-like” formulas [2, Chapter 10]. The even simpler BBP-formula
(A.14) appears in [3, eq. (18)] with two terms at a time, attributed to Knuth.
I note that the formulas in the Appendix emerged quite naturally as
accelerations of simple series, and all were derived by hand.
As we shall see, this acceleration method can also be used to transform
divergent series into convergent ones, in a process more properly called series
deceleration. For example, decelerating the divergent series
(6) π
33=
X
n=0
(3)n
2n+ 1
gives the convergent, fractional BBP-formula (A.18). This argument, of
course, is no proof, but I also give an alternative, rigorous demonstration.
The general principle behind these formulas is an acceleration scheme
that allows one to approximate an alternating, “sporadic” sum
S=a0ak+a2ka3k+·· ·
from a finite number of terms a0, a1, . . . , anof the complete sequence. The
general theory—in particular the idea of letting the aibe the moments of
a measure, writing Sas an integral with respect to that measure, and ap-
proximating the integrand by means of Chebyshev polynomials—is strongly
indebted to that in [7] which it generalizes from k= 1 to arbitrary k.
As another example of this method, we show how to compute numerically
the constant
(7) K=
X
n=0
(1)nζ(n+ 1/2)
2n+ 1 =2.1577829966 . . . ,
making the most of a small number of given integer and half-integer zeta-
values, ζ(n) and ζ(n+ 1/2). The constant Kis known as the Schnecken-
konstante and arises in connection with the Spiral of Theodorus [6].
Nilakantha’s accelerated series for π3
2. Alternating, sporadic series. Let µbe a finite, signed measure
on [0,1] with moments
(8) ai=
1
0
xidµ, i 0,
converging to zero for i→ ∞, and consider the alternating, sporadic series
(9) S=
1
0
1 + xk=
X
i=0
(1)iaik
for some integer k1. Let there be given a polynomial
(10) P(x) =
m+k1
X
i=0
bixi
with P(u) = 1 for uk=1, and write
(11) Q(x) = 1P(x)
1 + xk=
m1
X
i=0
cixi.
Define a new measure µ0with density P(x) with respect to µ, i.e., 0=
P(x), and moments
a0
i=
1
0
xi0=
m+k1
X
j=0
bjai+j,
and consider the transformed series
S0=
1
0
0
1 + xk=
X
i=0
(1)ia0
ik.
Write the difference between the old and the new series as
S=SS0=
1
0
Q(x)=
m1
X
i=0
ciai.
Repeating this process gives a sequence of measures µ(n)with densities
(n)=P(x)nand moments
(12) a(n)
i=
1
0
xi(n)
as well as a sequence of transformed series
(13) S(n)=
1
0
(n)
1 + xk=
X
i=0
(1)ia(n)
i
4 D. Brink
with differences
(14) S(n)=S(n)S(n+1) =
1
0
Q(x)(n)=
m1
X
i=0
cia(n)
i.
After nsteps one has
S=
n1
X
i=0 S(i)+S(n).
If Mis the maximum of |P(x)|on the interval [0,1], then (13) gives the
bound
(15) |S(n)| ≤ Mn
1
0
d|µ|
1 + xk,
cf. Remark 1 below. So if M < 1, one gets the accelerated series
(16) S=
X
n=0 S(n)
with convergence Mn, i.e., S(n)=O(Mn).
Remark 1.Asigned measure may take negative as well as positive
values. Even if µwere required to be positive, the transformed measure µ0
would still take negative values if the density function P(x) did. By Jordan’s
Decomposition Theorem [4], one can write µ=µ+µwith unique positive
measures µ+and µwith disjoint supports. The theory of integration with
respect to a signed measure can thus be reduced to that of a usual, positive
measure. Absolute integrals such as the one appearing in (15) are defined
by means of the total variation |µ|=µ++µ.
The assumption that the moments (8) converge to zero is equivalent to
µ({1}) = 0 and guarantees that
1
1 + xk= 1 xk+x2k− · · ·
can be integrated termwise, say, by Lebesgue’s Dominated Convergence The-
orem.
It is a result of Hausdorff [12, Satz I] that a sequence of real numbers
a0, a1, a2, . . . is the sequence of moments of a finite, positive measure on [0,1]
if and only if it is totally monotonic, i.e.,
(17) nai0 for all i, n 0.
As above, ai=aiai+1 denotes the negated forward difference operator.
Similarly, also by Hausdorff [12, Satz II], the aiare the moments of a finite,
Nilakantha’s accelerated series for π5
signed measure on [0,1] if and only if
(18) sup
n0
n
X
i=0
n
iniai
<.
The latter condition thus implies that (ai) is the difference between two
totally monotonic sequences. It is seen directly from the identity
n
X
i=0 n
iniai=a0
that (17) implies (18).
Note that the moments (8) need not be of the same sign, not even even-
tually, so that in reality the series (9) is not necessarily alternating.
For later use we also note that ai= 1/(i+ 1) and a
i= 1/(2i+ 1) are the
moments of the usual Lebesgue measure µand the measure µwith density
=dµ/2x, respectively.
Remark 2.In order to compute Snumerically, we approximate it by
the first difference S. To minimize the error S0, we have to choose P(x)
as a polynomial of high degree that approximates zero uniformly on [0,1].
In light of (11), this suggests taking
P(x)=1(1 + xk)Q(x),
where the polynomial Q(x) is a Chebyshev approximation to 1/(1 + xk).
This will be carried out in more detail in Sections 4 and 6.
Remark 3.On the other hand, if we wish to transform Sinto an exact,
accelerated series (16), we have to choose P(x) as a simple polynomial of low
degree, so that the terms (14) can be computed explicitly. If, for example,
P(x) is of the form xj(1 x)n, then the terms of the transformed series S0
are a0
i=nai+j.
The binomial sum
n
X
j=0 n
j(1)j
x+j=n!
x(x+ 1) · ·· (x+n)
is well known. It holds as an identity in Q(x) and can be easily shown as a
partial fraction decomposition. The substitution x=i+ 1 thus gives
(19) nai=n!i!
(n+i+ 1)!
for the sequence ai= 1/(i+ 1). Similarly, letting x=i+ 1/2 gives
(20) na
i=4nn!(2i)!(n+i)!
i!(2n+ 2i+ 1)!
for a
i= 1/(2i+ 1). This will be of use later.
6 D. Brink
3. The Nilakantha transform. Let k= 2 throughout this section.
Then P(x) must satisfy P(±i) = 1. We can take for P(x) any product of
x2,x(1 x)2
2,(1 x)4
4.
Let
(21) P(x) = x(1 x)2
2,
so that
Q(x) = 1 x
2.
The transformed series S0has terms
a0
i=2ai+1
2=ai+1 2ai+2 +ai+3
2,
hence the first step of the transformation becomes
(22) S=a0a1
2+
X
i=0
ai+1 2ai+2 +ai+3
2.
The ntimes transformed series S(n)has terms
a(n)
i=1
2n2nai+n,
so that we get the Nilakantha transform
(23) S=
X
n=0
1
2n2nanan+1
2
with convergence 13.5n, since
(24) M=P(1/3) = 2/27.
Example 4.To accelerate the Gregory–Leibniz series (1), we let ai=
1/(i+ 1). The first step (22) of the transformation is precisely Nilakantha’s
series (2). To compute the fully accelerated series (23), we use the iden-
tity (19) and get (A.1).
Example 5.Taking instead P(x) as
(25) (1 x)4
4,x3(1 x)2
2,x4(1 x)4
4
gives the three series (A.2), (A.3), (A.4) with M= 1/4, 54/3125, 1/1024,
respectively.
Example 6.The Newton–Euler series (5) can be rewritten as
π
22=
X
n=0
(1)n
4n+ 1 +
X
n=0
(1)n
4n+ 3
Nilakantha’s accelerated series for π7
with two alternating, sporadic series corresponding to a
i= 1/(2i+ 1) and
a∗∗
i= 1/(2i+ 3). Accelerating each series separately using
P(x) = x(1 x)2
2,(1 x)4
4,x3(1 x)2
2
and adding the results gives (A.5), (A.6), (A.7), respectively.
Remark 7.It follows in advance from (24) that (A.1) and (A.5) con-
verge as 13.5n, but this is also evident from the expressions themselves,
by Stirling’s Formula. A similar remark applies to all other formulas in the
Appendix.
Remark 8.The factors
(2n)!(2n)!(3n)!
n!(6n)! ,(2n)!(5n)!(6n)!
(3n)!(10n)!
appearing in (A.5) and (A.7) happen to be reciprocal integers by a criterion
of Landau (1900), anticipated by Chebyshev (1852) and Catalan (1874).
Such expressions are not too common and have been completely classified
(in a suitable sense) [5].
Remark 9.I stress that there is no evidence that Nilakantha derived
(2) the way we have done here, much less that he knew (3). Of course, the
transformation (22) is straightforward to verify directly, making (1) and (2)
essentially equivalent.
4. Numerical approximations. Let k1 be given. The Chebyshev
polynomials of the first kind, Tm(x), are given recursively by T0(x) = 1,
T1(x) = xand
Tm(x) = 2xTm1(x)Tm2(x).
The zeros of Tm(x) are
ηi= cos (2i1)π
2m, i = 1, . . . , m.
Let Q(x) be the Chebyshev approximation of order mof 1/(1 + xk) on
the interval [0,1], i.e., Q(x) is the polynomial of degree less than magreeing
with 1/(1 + xk) at the mpoints (1 + ηi)/2. Since (1 + ηi)/2 are the zeros of
Tm(1 2x), Q(x) satisfies
Q(x)1
1 + xkmodulo Tm(1 2x)
and can be computed from this congruence by the Euclidean Algorithm.
Thus, P(x) will be the polynomial of degree less than m+kwith zeros
(1 + ηi)/2 and P(u) = 1 for uk=1. Lagrange interpolation gives the
8 D. Brink
explicit expression
P(x) = Tm(1 2x)X
uk=1
u
βmk·1 + xk
ux
with βm=Tm(1 2u).
In order to evaluate the maximum Mof |P(x)|for 0 x1 as m→ ∞,
first note |Tm(12x)| ≤ 1. For a fixed u, the numbers βmsatisfy β0= 1, β1=
12uand the recursion βm= (24u)βm1βm2.Hence, βm= (λm
1+λm
2)/2
with the roots λiof the characteristic polynomial λ2(2 4u)λ+ 1.We
may suppose |λ1|>|λ2|, and conclude that βm∼ |λ1|m/2.Finally, let λbe
the minimum of |λ1|as uruns through the roots of unity with uk=1.
Then M=O(λm) as m→ ∞.
Some values of λare given in Table 1. Note that the value λ= 5.828 for
k= 1 was found in [7].
Table 1. Values of λ
k12345678910
λ5.828 4.612 3.732 3.220 2.890 2.659 2.488 2.356 2.250 2.164
Example 10.Suppose we want to compute numerically the alternating,
sporadic sum
S=
1
0
1 + x2=
X
i=0
(1)ia2i,
and that we have at our disposal the terms a0, a1, a2, . . . , a99. Letting k= 1
and m= 50 and using only every second term, a0, a2, a4, . . . , a98, we expect
a relative error of 5.82850, or 38 correct, significant digits of S. Letting
k= 2 and m= 100, using all 100 available terms, we expect an error of
4.612100, or 66 correct digits.
Consider the constant (7), and rewrite it as
K=ζ1
2
X
i=0
(1)iζ(i+ 3/2)
2i+ 3
in order to bypass the singularity at z= 1. Suppose that the zeta-values
ζ(i/2+3/2) are available for i= 0,1,...,99. Using the first method gives
42 digits of S, while the second method gives 70 digits, in agreement with
our expectations. The second method can be carried out in Pari as follows:
T=subst(poltchebi(100),x,1-2*x)
Q=lift(1/Mod(1+x^2,T))
c(i)=polcoeff(Q,i)
a(i)=zeta(i/2+3/2)/(i+3)
K=zeta(1/2)-sum(i=0,99,c(i)*a(i))
We return to this example in Section 6.
Nilakantha’s accelerated series for π9
Remark 11.In the above example, it was assumed that the terms a0, a1,
...,a99 were simply given in advance. In practice, one might obviously have
to compute them first. Using k= 2 has the advantage that the integer
zeta-values ζ(n) are much faster to compute than the half-integer values
ζ(n+ 1/2). On the other hand, these values then need to be computed to a
precision of up to 10 extra digits due to the numerically larger coefficients ci.
Also, the cican be computed particularly efficiently for k= 1 (cf. [7]).
Remark 12.For k= 2, we can compare the (optimal) value λ= 4.612
obtained from Chebyshev polynomials with the values λ=M1/deg Pfrom
the polynomials P(x) given in Section 3. Of these, the Nilakantha trans-
formation (21) has the best convergence, i.e., λ= 2.381. The other three
transformations (25) have λ= 1.414, 2.252, 2.378, respectively.
5. Geometrically converging series. Let µbe a finite, signed mea-
sure on [0,1] with arbitrary moments (8), and consider the alternating, ge-
ometrically converging series
(26) S=
1
0
1 + θxk=
X
i=0
(θ)iaki
for k1 and 0 < θ < 1. Let P(x) be given as in (10) with P(u/θ1/k) = 1
for uk=1, and write
Q(x) = 1P(x)
1 + θxk.
As before, we define a sequence of measures µ(n)with (n)=P(x)nand
moments (12), and we get a sequence of transformed series
S(n)=
1
0
(n)
1 + θxk
with differences (14) as well as an accelerated series (16) with S(n)=
O(Mn), where Mis the maximum of |P(x)|on [0,1].
Example 13.The arcus tangent series
(27) arctan θ
θ=
X
i=0
(θ)i
2i+ 1
has the form (26) with k= 1 and ai= 1/(2i+ 1). To accelerate it, P(x)
must satisfy P(1) = 1, and we can take any product of
θx, θ(1 x)
θ+ 1 .
10 D. Brink
Letting
(28) P(x) = θ(1 x)
θ+ 1
and using (20) gives Euler’s accelerated series
(29) arctan θ
θ=1
θ+ 1
X
n=04θ
θ+ 1nn!n!
(2n+ 1)!
with M=θ/(θ+ 1).
Note that the original series (27) converges for |θ|<1, whereas the
accelerated series (29) converges for |θ|<|θ+ 1|, or Re(θ)>1/2. The pre-
ceding discussion shows that the two series agree for 0 < θ < 1. The Identity
Theorem for holomorphic functions and the fact that uniform convergence
preserves holomorphicity show that (29) holds for Re(θ)>1/2.
Inserting θ= 1, 1/3, 3 gives three classical formulas such as (4).
Also note that (4) is the Euler transform of the Gregory–Leibniz series,
i.e., the acceleration corresponding to the negated difference operator , or
P(x) = 1x
2.
Historical note 14.Euler develops the Euler transformation and de-
rives (4) and (29) from (1) and (27) in his Institutiones Calculi Differentialis
[9, Part II, Chapter 1] from 1755. Much earlier, he had given a series for
arcsin2xessentially equivalent to (29) in a letter to Johann Bernoulli dated
10 December 1737 [15]. Euler proves (29) again (twice) as well as the Machin-
like formula π= 20 arctan 1/7 + 8 arctan 3/79, and computes the two terms
with 13 and 17 correct decimals, respectively, but without adding them, in
1779 [10]. He extends this calculation and computes 21 correct decimals of
πin [11]. Several sources on the chronology of πstate that Euler did this
calculation in 1755 and/or in less than an hour. It seems from the above,
though, that the calculation could not have been carried out before 1779.
Regarding the duration, the relevant passage reads: “totusque hic calculus
laborem unius circiter horae consum[p]sit” (and the entire calculation took
about an hour’s work).
Example 15.Taking
P(x) = θ2x(1 x)
θ+ 1
rather than (28) gives the accelerated series
arctan θ
θ=1
4(θ+ 1)
X
n=04θ2
θ+ 1n4n
2n13θ+ 4
4n+ 1 θ
4n+ 3
with M=θ2/4(θ+ 1).
Nilakantha’s accelerated series for π11
By the same argument as before, this formula holds on a complex domain
bounded by the curve |θ|2= 4|θ+ 1|, or
(x2+y2)2= 16((x+ 1)2+y2)
in real variables. This quartic, algebraic curve is a lima¸con of Pascal, named
after ´
Etienne Pascal, the father of Blaise Pascal, and first studied in 1525
by Albrecht D¨urer [13] (1).
Fig. 1. Lima¸con of Pascal
Inserting θ= 1, 1/3, 3 gives (A.8), (A.9), (A.10).
Note that the small loop around –1 is not included in the domain of
convergence, corresponding nicely to the fact that arctan has a singularity
at ±i.
Example 16.Letting
P(x) = θ3x(1 x)2
(θ+ 1)2
gives the formidable expression
arctan θ
θ=1
9(θ+ 1)2
X
n=016θ3
(θ+ 1)2n
×(2n)!(2n)!(3n)!
n!(6n)! 5θ2+ 15θ+ 9
6n+ 1 θ2
6n+ 5
with M= 4θ3/27(θ+ 1)2.
The domain of convergence is bounded by the sextic, lima¸con-like curve
16(x2+y2)3= 729((x+ 1)2+y2)2.
Inserting θ= 1, 1/3, 3 gives (A.11), (A.12), (A.13).
(1) I am grateful to my friend Kasper K. S. Andersen for identifying this curve.
12 D. Brink
Formulas (A.8) and (A.11) are examples of van Wijngaarden’s trans-
formation [7], i.e., they are the accelerations of the Gregory–Leibniz series
corresponding to the polynomials
P(x) = x(1 x)
2,x(1 x)2
4.
Example 17.For k= 2 and ai= 1/(i+1), the general arctan series (27)
cannot be accelerated as in the previous examples. It may, however, for
specific choices of θ. Let θ= 1/3. Then we must have P(±i3) and can
take any product of
x2
3,(1 x)3
8.
Letting
P(x) = (1 x)3
8,x2(1 x)3
24
gives the series (A.14), (A.15) with M= 1/8, 9/6250, respectively.
Example 18.The convergence of the accelerated series (16), and its
identity with (26), was proved under Hausdorff’s condition (18) and 0 <
θ < 1. It is a common phenomenon, however, that acceleration techniques
work in more general settings and even for divergent series [7, Remark 6].
Consider the divergent series (6), obtained by inserting θ= 3 into (27). Let
k= 2 and θ= 3. Then P(x) must satisfy
P±i
3= 1.
We can take for P(x) any product of
3x2,9x(1 x)3
8,27(1 x)6
64 .
Letting P(x) be
9x(1 x)3
8,27x3(1 x)3
8,27(1 x)6
64
gives the three series (A.16), (A.17), (A.18) with M= 243/2048, 27/512,
27/64, respectively.
These formulas can be checked numerically to many digits, but of course
the above argument is no proof (although I like to think that Euler would
have appreciated it).
Remark 19.A quick, rigorous proof of (A.18) could go as follows. Write
Mercator’s Formula with six terms at a time,
log(1 z) =
X
n=0
z6nz
6n+ 1 +· ·· +z6
6n+ 6.
Insert z=eiπ/63/2 and take imaginary parts to get (A.18), q.e.d.
Nilakantha’s accelerated series for π13
Similar proofs of (A.2) and (A.14) are possible: Insert z= (1 + i)/2 and
z=eiπ/3/2 into Mercator’s Formula with four and three terms at a time,
respectively.
6. Numerical approximations again. To approximate the geomet-
rically converging series (26) numerically, let k1 be given, but now let
Q(x) agree with 1/(1 + θxk) at the points (1 + ηi)/2, i.e.,
Q(x)1
1 + θxkmodulo Tm(1 2x).
Then
P(x) = Tm(1 2x)X
uk=1
u
βmk·1 + θxk
uθ1/kx
with βm=Tm(1 21/k ). Again, βm∼ |λ1|m/2 with λ1the numerically
greater root of the characteristic polynomial λ2(2 41/k )λ+ 1. We
conclude that M=O(λm) as m→ ∞ with
λ= min{|λ1|:uk=1}.
Table 2 gives λ=λθfor various values of kand θ.
Table 2. Values of λθ
k λ1/2λ1/3λ1/4λ1/5λ1/6
1 9.899 13.93 17.94 21.95 25.96
2 6.129 7.328 8.352 9.263 10.09
3 4.607 5.254 5.782 6.236 6.636
4 3.829 4.264 4.612 4.905 5.160
5 3.357 3.685 3.942 4.157 4.343
6 3.040 3.303 3.508 3.678 3.823
7 2.811 3.031 3.202 3.342 3.462
8 2.636 2.827 2.973 3.094 3.196
9 2.499 2.667 2.796 2.902 2.991
10 2.388 2.539 2.654 2.748 2.828
Example 20.We return to the computation of the constant Kfrom
Example 10. Write
K=π
41 + ζ1
2
X
i=0
(1)iζ(i+ 3/2) 1
2i+ 3
to get a geometrically converging series, with θ= 1/2. Suppose again the
14 D. Brink
zeta-values ζ(i/2 + 3/2) are given for i= 0,1,...,99. Using only the terms
a0, a2, a4, . . . , a98, we expect an error of 9.89950, or 50 correct digits (cf. Ta-
ble 2). Using all 100 available terms, we expect an error of 6.129100, or 79
correct digits. In practice, the two methods give 53 and 82 digits, respec-
tively, confirming the theory. The second method can be carried out in Pari
as follows:
T=subst(poltchebi(100),x,1-2*x)
Q=lift(1/Mod(1+x^2/2,T))
c(i)=polcoeff(Q,i)
a(i)=2^(i/2)*(zeta(i/2+3/2)-1)/(i+3)
K=Pi/4-1+zeta(1/2)-sum(i=0,99,c(i)*a(i))
Appendix. Series for π
(A.1) 3π
2=
X
n=0
1
2n3n
n14
3n+ 1 +1
3n+ 2,
(A.2) π=
X
n=01
4n2
4n+ 1 +2
4n+ 2 +1
4n+ 3,
(A.3) 125π
2=
X
n=01
2n5n
2n1
×208
5n+ 1 22
5n+ 2 +8
5n+ 3 7
5n+ 4,
(A.4) 1024π=
X
n=01
4n8n
4n1
×3183
8n+ 1 +117
8n+ 3 15
8n+ 5 5
8n+ 7,
(A.5) 9π
2=
X
n=0
8n(2n)!(2n)!(3n)!
n!(6n)! 19
6n+ 1 +1
6n+ 5,
(A.6) 162π=
X
n=0
(64)n8n
4n1
×75
8n+ 1 +13
8n+ 3 3
8n+ 5 5
8n+ 7,
(A.7) 625π
2=
X
n=0
(8)n(2n)!(5n)!(6n)!
(3n)!(10n)!
×1339
10n+ 1 +184
10n+ 3 16
10n+ 7 11
10n+ 9,
Nilakantha’s accelerated series for π15
(A.8) 2π=
X
n=0
(2)n4n
2n17
4n+ 1 1
4n+ 3,
(A.9) 8π
3=
X
n=01
3n4n
2n115
4n+ 1 1
4n+ 3,
(A.10) 16π
33=
X
n=0
(9)n4n
2n113
4n+ 1 3
4n+ 3,
(A.11) 9π=
X
n=0
(4)n(2n)!(2n)!(3n)!
n!(6n)! 29
6n+ 1 1
6n+ 5,
(A.12) 243π=
X
n=01
3n(2n)!(2n)!(3n)!
n!(6n)! 131
6n+ 1 1
6n+ 5,
(A.13) 16π
33=
X
n=0
(27)n(2n)!(2n)!(3n)!
n!(6n)! 11
6n+ 1 1
6n+ 5,
(A.14) 4π
33=
X
n=01
8n2
3n+ 1 +1
3n+ 2,
(A.15) 500π
3=
X
n=0
1
24n5n
2n1872
5n+ 1 +57
5n+ 2 +12
5n+ 3 +7
5n+ 4 ,
(A.16) 256π
33=
X
n=09
8n4n
n1103
4n+ 1 +72
4n+ 2 +15
4n+ 3,
(A.17) 1024π
33=
X
n=027
8n6n
3n1637
6n+ 1 +6
6n+ 3 27
6n+ 5,
(A.18) 64π
33=
X
n=027
64 n
×16
6n+ 1 +24
6n+ 2 +24
6n+ 3 +18
6n+ 4 +9
6n+ 5.
References
[1] V. Adamchik and S. Wagon, A simple formula for π, Amer. Math. Monthly 104
(1997), 852–855.
[2] J. Arndt and C. Haenel, Pi—Unleashed, 2nd ed., Springer, Berlin, 2001.
[3] D. H. Bailey, A compendium of BBP-type formulas for mathematical constants,
2013; www.davidhbailey.com/dhbpapers/bbp-formulas.pdf.
[4] P. Billingsley, Probability and Measure, 3rd ed., Wiley, New York, 1995.
[5] J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions,
J. London Math. Soc. 79 (2009), 422–444.
16 D. Brink
[6] D. Brink, The spiral of Theodorus and sums of zeta-values at the half-integers, Amer.
Math. Monthly 119 (2012), 779–786.
[7] H. Cohen, F. Rodriguez Villegas and D. Zagier, Convergence acceleration of alter-
nating series, Experiment. Math. 9 (2000), 3–12.
[8] L. Euler, De summis serierum reciprocarum, Comment. Acad. Sci. Petropol. 7
(1740), 123–134; online: eulerarchive.maa.org, Enestr¨om index E41.
[9] L. Euler, Institutiones Calculi Differentialis. . . , St. Petersburg, 1755; [E212].
[10] L. Euler, Investigatio quarundam serierum, quae ad rationem peripheriae circuli ad
diametrum vero proxime definiendam maxime sunt accommodatae, Nova Acta Acad.
Sci. Imp. Petropol. 11 (1798), 133–149; [E705].
[11] L. Euler, Series maxime idoneae pro circuli quadratura proxime invenienda, in:
Opera Postuma I, St. Petersburg, 1862, 288–298; [E809].
[12] F. Hausdorff, Momentprobleme f¨ur ein endliches Intervall, Math. Z. 16 (1923), 220–
248.
[13] J. D. Lawrence, A Catalog of Special Plane Curves, Dover, New York, 1972.
[14] R. Roy, The discovery of the series formula for πby Leibniz, Gregory and Nilakan-
tha, Math. Mag. 63 (1990), 291–306.
[15] P. St¨ackel, Eine vergessene Abhandlung Leonhard Eulers ¨uber die Summe der re-
ziproken Quadrate der nat¨urlichen Zahlen, Bibliotheca Math. 8 (1908), 37–60.
David Brink
Akamai Technologies
Larslejsstræde 6
1451 København K, Denmark
E-mail: dbrink@akamai.com
Received on 28.10.2014
and in revised form on 22.8.2015 (7975)
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