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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)!

2n−1(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 13.5−n, whereas the Euler transform converges as 2−n. 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

2√2=

∞

X

n=0

(−1)n(n−1)/2

2n+ 1

gives the accelerated series (A.5) (see the Appendix), also with convergence

13.5−n. 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

3√3=

∞

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=a0−ak+a2k−a3k+·· ·

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

dµ

1 + xk=

∞

X

i=0

(−1)iaik

for some integer k≥1. Let there be given a polynomial

(10) P(x) =

m+k−1

X

i=0

bixi

with P(u) = 1 for uk=−1, and write

(11) Q(x) = 1−P(x)

1 + xk=

m−1

X

i=0

cixi.

Deﬁne a new measure µ0with density P(x) with respect to µ, i.e., dµ0=

P(x)dµ, and moments

a0

i=

1

0

xidµ0=

m+k−1

X

j=0

bjai+j,

and consider the transformed series

S0=

1

0

dµ0

1 + xk=

∞

X

i=0

(−1)ia0

ik.

Write the diﬀerence between the old and the new series as

∇S=S−S0=

1

0

Q(x)dµ =

m−1

X

i=0

ciai.

Repeating this process gives a sequence of measures µ(n)with densities

dµ(n)=P(x)ndµ and moments

(12) a(n)

i=

1

0

xidµ(n)

as well as a sequence of transformed series

(13) S(n)=

1

0

dµ(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)dµ(n)=

m−1

X

i=0

cia(n)

i.

After nsteps one has

S=

n−1

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+−µ−with unique positive

measures µ+and µ−with 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) ∇nai≥0 for all i, n ≥0.

As above, ∇ai=ai−ai+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

n≥0

n

X

i=0

n

i∇n−iai

<∞.

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

i∇n−iai=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 µ∗with density

dµ∗=dµ/2√x, 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=a0−a1

2+

∞

X

i=0

ai+1 −2ai+2 +ai+3

2.

The ntimes transformed series S(n)has terms

a(n)

i=1

2n∇2nai+n,

so that we get the Nilakantha transform

(23) S=

∞

X

n=0

1

2n∇2nan−an+1

2

with convergence 13.5−n, 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

2√2=

∞

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 13.5−n, 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 k≥1 be given. The Chebyshev

polynomials of the ﬁrst kind, Tm(x), are given recursively by T0(x) = 1,

T1(x) = xand

Tm(x) = 2xTm−1(x)−Tm−2(x).

The zeros of Tm(x) are

ηi= cos (2i−1)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

u−x

with βm=Tm(1 −2u).

In order to evaluate the maximum Mof |P(x)|for 0 ≤x≤1 as m→ ∞,

ﬁrst note |Tm(1−2x)| ≤ 1. For a ﬁxed u, the numbers βmsatisfy β0= 1, β1=

1−2uand the recursion βm= (2−4u)βm−1−βm−2.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(λ−m) 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

dµ

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 5.828−50, or 38 correct, signiﬁcant digits of S. Letting

k= 2 and m= 100, using all 100 available terms, we expect an error of

4.612−100, 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=M−1/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

dµ

1 + θxk=

∞

X

i=0

(−θ)iaki

for k≥1 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) = 1−P(x)

1 + θxk.

As before, we deﬁne a sequence of measures µ(n)with dµ(n)=P(x)ndµ and

moments (12), and we get a sequence of transformed series

S(n)=

1

0

dµ(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/θ) = 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) = 1−x

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=0−4θ2

θ+ 1n4n

2n−13θ+ 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=0−16θ3

(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(±i√3) 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/6√3/2 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 k≥1 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

u−θ1/kx

with βm=Tm(1 −2uθ−1/k ). Again, βm∼ |λ1|m/2 with λ1the numerically

greater root of the characteristic polynomial λ2−(2 −4uθ−1/k )λ+ 1. We

conclude that M=O(λ−m) 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

4−1 + 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 9.899−50, or 50 correct digits (cf. Ta-

ble 2). Using all 100 available terms, we expect an error of 6.129−100, 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

n−14

3n+ 1 +1

3n+ 2,

(A.2) π=

∞

X

n=0−1

4n2

4n+ 1 +2

4n+ 2 +1

4n+ 3,

(A.3) 125π

2=

∞

X

n=0−1

2n5n

2n−1

×208

5n+ 1 −22

5n+ 2 +8

5n+ 3 −7

5n+ 4,

(A.4) 1024π=

∞

X

n=0−1

4n8n

4n−1

×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) 16√2π=

∞

X

n=0

(−64)n8n

4n−1

×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

2n−17

4n+ 1 −1

4n+ 3,

(A.9) 8π

√3=

∞

X

n=0−1

3n4n

2n−115

4n+ 1 −1

4n+ 3,

(A.10) 16π

3√3=

∞

X

n=0

(−9)n4n

2n−113

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) 24√3π=

∞

X

n=0−1

3n(2n)!(2n)!(3n)!

n!(6n)! 131

6n+ 1 −1

6n+ 5,

(A.13) 16π

3√3=

∞

X

n=0

(−27)n(2n)!(2n)!(3n)!

n!(6n)! 11

6n+ 1 −1

6n+ 5,

(A.14) 4π

3√3=

∞

X

n=0−1

8n2

3n+ 1 +1

3n+ 2,

(A.15) 500π

√3=

∞

X

n=0

1

24n5n

2n−1872

5n+ 1 +57

5n+ 2 +12

5n+ 3 +7

5n+ 4 ,

(A.16) 256π

3√3=

∞

X

n=09

8n4n

n−1103

4n+ 1 +72

4n+ 2 +15

4n+ 3,

(A.17) 1024π

3√3=

∞

X

n=0−27

8n6n

3n−1637

6n+ 1 +6

6n+ 3 −27

6n+ 5,

(A.18) 64π

3√3=

∞

X

n=0−27

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)