Interesting Ramanujan-Like Series Associated with Powers of Central Binomial Coefficients

Hongwei Chen Department of Mathematics Christopher Newport University Newport News, VA 23606 USA

                         hchen@cnu.edu

Abstract We find various Ramanujan-like series that involve the central binomial coefficients. In contrast with the traditional hypergeometric function approach, our method uses Fourier-Legendre series expansions via specialization, inner product, and Parseval’s identity. Several classical identities are recovered as particular cases.

1 Introduction

A Ramanujan-like series for 1/π is an identity of the following form:

∑_{n=0}^{∞} \frac{an + b}{c^{n}} s(n) = \frac{1}{π} ,

where a and b are algebraic numbers, c ∈ ℕ, and s(n) is an integer sequence obeying a certain recurrence relation. More than 100 years ago, without proof, Ramanujan recorded 17 formulas for 1/π at the end of his first paper published in England [13]. The first two formulas (with slight modification), which surprisingly appeared in the Disney movie High School Musical, are

∑_{n=0}^{∞} \frac{6n + 1}{16^{2n}} \binom{2n}{n}^{3} = \frac{4}{π}  (1)

and

∑_{n=0}^{∞} \frac{42n + 5}{64^{2n}} \binom{2n}{n}^{3} = \frac{16}{π},  (2)

where \binom{2n}{n} is the central binomial coefficient defined by (2n)!/(n!)^{2}. Except for Chowla’s proof of (1) in 1928 [7], Ramanujan’s series were forgotten by the mathematical community until 1987. The Borwein brothers [3] finally succeeded in proving all 17 of Ramanujan’s series for 1/π via modular forms. Moreover, as an application, they used one modified Ramanujan series to calculate the digits of π and were able to obtain roughly 50 digits of π per term. Since 2002, Guillera et al. [10, 11] have discovered many new Ramanujan-like series for 1/π^{2}. An excellent survey on the work of Ramanujan-like series before 2009 can be found in [2].

Following the ideas of the Borweins and Guillera et al., exploring new classes of Ramanujanlike series has become an active area of contemporary research. In my previous work [6], the following Ramanujan-like series

∑_{n=1}^{∞} \frac{\binom{2n}{n}^{2}H_{n}}{16^{n}(2n − 1)^{2}} = \frac{12 − 16 ln 2}{π},

where H_{n} denotes the nth harmonic number, was established by applying a differential operator to a hypergeometric function _{2}F_{1}-identity. Along this line, using the Gauss summation theorem and an extended_{3}F_{2}-series of Watson and Whipple type, Wang and Chu [16] offered a systematic evaluation of series like

∑_{n=1}^{∞} \frac{\binom{2n}{n}^{2}H_{n}}{16^{n}p(n)} and ∑_{n=1}^{∞} \frac{\binom{2n}{n}^{2}H_{n}^{(2)}}{16^{n}p(n)} ,

where p(n) = n + λ, 1 + 2n − 2λ or (1 + 2n − 2λ)^{2} for λ ∈ ℕ, and H_{n}^{(2)} = ∑_{k=1}^{n} 1/k^{2}. In particular, they obtained the following interesting Ramanujan-like series:

∑_{n=1}^{∞} \frac{\binom{2n}{n}^{2}H_{n}^{(2)}}{16^{n}(2n − 1)} = 4 − \frac{π}{3} − \frac{8}{π};

∑_{n=1}^{∞} \frac{\binom{2n}{n}^{2}H_{n}^{(2)}}{16^{n}(2n − 1)^{2}} = −12 + \frac{2π}{3} + \frac{32}{π}.

The aim of the present paper is to study the Ramanujan-like series involving powers of central binomial coefficients. In contrast to [4, 16], our approach is based on Fourier-Legendre series expansions. We will derive a variety of Ramanujan-like series including

∑_{n=0}^{∞} \frac{(−1)^{n}}{16^{n}} \binom{2n}{n}^{2} = \frac{\sqrt{2} Γ^{2}(1/4)}{4π^{3/2}};  (3)

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{64^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{3} = \frac{8}{π};  (4)

∑_{n=0}^{∞} \frac{(4n + 1)}{256^{n}(n + 1)^{2}(1/2 − n)^{2}} \binom{2n}{n}^{4} = \frac{128}{3π^{2}}.  (5)

Here Γ(x) indicates the gamma function.

The rest of paper is organized as follows. In Section 2, three combinatorial identities are established via the Wilf-Zeilberger method. These identities are used to derive several Fourier-Legendre series expansions. By using specialization, inner product, and Parseval’s identity, those series expansions are used to find various Ramanujan-like series in Section 3. The paper ends with two remarks. To ensure accuracy, all formulas appearing in this paper were verified numerically by Mathematica.

2 Some Fourier-Legendre series expansions

Recall the Legendre polynomial [9, pp. 983–985]

P_{n}(x) = \frac{1}{2^{n} n!} \frac{d^{n}}{dx^{n}} (x^{2} − 1)^{n}.

It is well-known that

∫_{−1}^{1} P_{n}(x)P_{m}(x) dx = \frac{2}{2n + 1} δ_{mn},

where δ_{mn} denotes the Kronecker delta. This implies that (P_{n}(x))_{n=0}^{∞} are orthogonal on [−1, 1]. Now, we define

P̃_{n}(x) = P_{n}(2x − 1).

Since x → 2x − 1 is an affine transformation that bijectively maps [0, 1] to [−1, 1], we see that (P̃_{n}(x))_{n=0}^{∞} are orthogonal on [0, 1] with

∫_{0}^{1} P̃_{n}(x)P̃_{m}(x) dx = \frac{1}{2n + 1} δ_{mn}.

Moreover, we have

P̃_{n}(x) = \frac{1}{n!} \frac{d^{n}}{dx^{n}} (x^{2} − x)^{n} = (−1)^{n} ∑_{k=0}^{n} \binom{n}{k}\binom{n + k}{k}(−x)^{k}.  (6)

Let f(x) ∈ L[0, 1]. The Fourier-Legendre series is defined by

f(x) = ∑_{n=0}^{∞} a_{n}P̃_{n}(x),

where

a_{n} = (2n + 1) ∫_{0}^{1} f(x)P̃_{n}(x) dx.  (7)

The following three lemmas display some features of the coefficients of variant Fourier-Legendre series.

Lemma 1. Let f(x) ∈ L[0, 1]. If f(x) = ∑_{n=0}^{∞} a_{n}P̃_{n}(x), then

f(1 − x) = ∑_{n=0}^{∞} (−1)^{n}a_{n}P̃_{n}(x).

Proof. Let a_{n}^{∗} be the nth coefficient of the Fourier-Legendre series for f(1 − x). By (7), we have

a_{n}^{∗} = (2n + 1) ∫_{0}^{1} f(1 − x)P̃_{n}(x) dx
= (2n + 1) ∫_{0}^{1} f(t)P̃_{n}(1 − t) dt (use t = 1 − x)
= (2n + 1) ∫_{0}^{1} f(t)(−1)^{n}P̃_{n}(t) dt = (−1)^{n}a_{n},

where we have used the fact that P̃_{n}(1−t) = P_{n}(−(2t−1)) = (−1)^{n}P_{n}(2t−1) = (−1)^{n}P̃_{n}(t).

Lemma 2. Let f(x) ∈ L[0, 1]. If f(x) = f(1 − x) for x ∈ [0, 1] and f(x) = ∑_{n=0}^{∞} a_{n}P̃_{n}(x), then a_{n} = 0 when n is odd.

Proof. Using (7), we rewrite a_{n} as

a_{n} = (2n + 1) (∫_{0}^{1/2} f(x)P̃_{n}(x) dx + ∫_{1/2}^{1} f(x)P̃_{n}(x) dx)

= (2n + 1) (∫_{0}^{1/2} f(x)P̃_{n}(x) dx + ∫_{0}^{1/2} f(1 − t)P̃_{n}(1 − t) dt) (use t = 1 − x)

= (2n + 1) ∫_{0}^{1/2} (1 + (−1)^{n})f(x)P̃_{n}(x) dx (use f(t) = f(1 − t)).

This leads to a_{n} = 0 when n is odd.

Recall Bonnet’s recursion formula [9, Formula 8.914 No.1, p. 985]

(n + 1)P_{n+1}(x) = (2n + 1)xP_{n}(x) − nP_{n−1}(x).

This implies that

(2n + 1)xP̃_{n}(x) = \frac{n + 1}{2}P̃_{n+1}(x) + \frac{2n + 1}{2}P̃_{n}(x) + \frac{n}{2}P̃_{n−1}(x),

which yields the following result.

Lemma 3. Let f(x) ∈ L[0, 1] and f(x) = ∑_{n=0}^{∞} a_{n}P̃_{n}(x). If

xf(x) = ∑_{n=0}^{∞} b_{n}P̃_{n}(x),

then for all n ≥ 1,

b_{n} = \frac{n + 1}{2(2n + 3)} a_{n+1} + \frac{1}{2} a_{n} + \frac{n}{2(2n − 1)} a_{n−1}.  (8)

The following two lemmas establish some identities that we will need later to simplify the coefficients of the Fourier-Legendre series.

Lemma 4. For any non-negative integer n,

∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k}\binom{2k}{k}4^{2n−k} = \binom{2n}{n}^{2}.  (9)

This identity (9) appears as (6.35) in [8], and is credited to E. T. Bell without providing a proof. For completeness, following Tauraso’s suggestion [14], we demonstrate a proof based on the Wilf-Zeilberger method (WZ-method).

Proof. Rewrite (9) as

∑_{k=0}^{2n} \frac{(−1)^{k}\binom{2n}{k}\binom{2n+k}{k}\binom{2k}{k}4^{2n−k}}{\binom{2n}{n}2} = 1.  (10)

Let

F (n, k) = \frac{(−1)^{k}\binom{2n}{k}\binom{2n+k}{k}\binom{2k}{k}4^{2n−k}}{\binom{2n}{n}2} .

Using the WZ-method, we find

G(n, k) = −\frac{2(4n + 3)k^{3}}{(2n − k + 1)(2n − k + 2)(2n + 1)^{2}} F (n, k).

Now it suffices to check

F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k).  (11)

Notice that

F (n + 1, k) = \frac{4(2n + k + 2)(2n + k + 1)(n + 1)^{2}}{(2n − k + 2)(2n − k + 1)(2n + 1)^{2}} F (n, k),
F (n, k + 1) = −\frac{(2k + 1)(2n + k + 1)(2n − k)}{2(k + 1)^{3}} F (n, k),
G(n, k + 1) = \frac{(4n + 3)(2k + 1)(2n + k + 1)}{(2n − k + 1)(2n + 1)^{2}} F (n, k);

with these (11) becomes the alleged identity

\frac{4(2n + k + 2)(2n + k + 1)(n + 1)^{2}}{(2n − k + 2)(2n − k + 1)(2n + 1)^{2}} − 1

= \frac{(4n + 3)(2k + 1)(2n + k + 1)}{(2n + 1 − k)(2n + 1)^{2}} + \frac{2(4n + 3)k^{3}}{(2n + 1 − k)(2n + 2 − k)(2n + 1)^{2}} ?

Combining the fractions shows that each side leads to

\frac{(4n + 3)(2 + 5k + k^{2} + 6n + 12kn + 4n^{2} + 8kn^{2})}{(2n − k + 1)(2n − k + 2)(2n + 1)^{2}}.

So (11) does actually hold. Since G is telescoping, we finally obtain

∑_{k=0}^{2n} F (n, k) = const.

This constant does not depend on n. Thus (10), and so (9), follows from setting n = 0.

Similarly, applying the WZ-method, we can establish Lemma 5. For any non-negative integer n,

∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k}\frac{3 · 7 ⋯ (4k − 1)}{(2k + 1)k!\binom{2k}{k}} = \frac{1}{(4n + 1)4^{n}} \binom{2n}{n};  (12)
∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k}\frac{5 · 9 ⋯ (4k + 1)}{(2k + 3)(2k + 1)k!\binom{2k}{k}} = \frac{1}{(4n + 3)(1 − 4n)4^{n}} \binom{2n}{n}.  (13)

Now we are ready to establish some Fourier-Legendre series in explicit form. Theorem 6. For x ∈ (0, 1),

\frac{1}{\sqrt{x(1 − x)}} = ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}} \binom{2n}{n}^{2}P̃_{2n}(x).  (14)

Proof. Let f(x) = 1/\sqrt{x(1 − x)}. Then f(x) = f(1 − x) for all x ∈ (0, 1). Using Lemma 2, a_{n} = 0 for odd n. Moreover,

a_{2n} = (4n + 1) ∫_{0}^{1} f(x)P̃_{2n}(x) dx.

Invoking (6) leads to

a_{2n} = (4n + 1) ∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k} ∫_{0}^{1} x^{k−1/2}(1 − x)^{−1/2} dx

= (4n + 1) ∑_{k=0}^{2n} \binom{2n}{k}\binom{2n + k}{k}(−1)^{k}B(k + 1/2, 1/2),

where B(x, y) is the Euler Beta function. Using

Γ(1/2) = \sqrt{π} and Γ(k + 1/2) = \frac{(2k)!}{4^{k} k!} \sqrt{π},  (15)

it follows that

B(k + 1/2, 1/2) = \frac{Γ(k + 1/2)Γ(1/2)}{Γ(k + 1)} = \frac{(2k)!}{4^{k} k!^{2}} π = \frac{π}{4^{k}} \binom{2k}{k}.

This, together with (9), implies

a_{2n} = (4n + 1)π ∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k}\binom{2k}{k}4^{−k} = \frac{(4n + 1)π}{16^{n}} \binom{2n}{n}^{2},

which proves (14).

Applying (8) to (14) gives

\frac{x}{\sqrt{x(1 − x)}} = ∑_{n=0}^{∞} \frac{(4n + 1)π}{2 · 16^{n}} \binom{2n}{n}^{2}P̃_{2n}(x)

+ ∑_{n=0}^{∞} \frac{(4n + 3)(2n + 1)π}{4 · 16^{n}(n + 1)} \binom{2n}{n}^{2}P̃_{2n+1}(x).  (16)

While applying (8) to (16) yields

\frac{x^{2}}{\sqrt{x(1 − x)}} = ∑_{n=0}^{∞} \frac{(4n + 1)(8n^{2} + 4n − 3)π}{8 · 16^{n}(n + 1)(2n − 1)} \binom{2n}{n}^{2}P̃_{2n}(x)

+ ∑_{n=0}^{∞} \frac{(4n + 3)(2n + 1)π}{4 · 16^{n}(n + 1)} \binom{2n}{n}^{2}P̃_{2n+1}(x).  (17)

Notice that

\sqrt{x(1 − x)} = \frac{x}{\sqrt{x(1 − x)}} − \frac{x^{2}}{\sqrt{x(1 − x)}}.

Combining (16) and (17) leads to Theorem 7. For x ∈ (0, 1),

\sqrt{x(1 − x)} = \frac{1}{8} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}(n + 1)(1 − 2n)} \binom{2n}{n}^{2}P̃_{2n}(x).  (18)

Applying the above process to (18) yields Theorem 8. For x ∈ (0, 1),

(\sqrt{x(1 − x)})^{3} = \frac{9}{8^{2}} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}(n + 1)(n + 2)(1 − 2n)(3 − 2n)} \binom{2n}{n}^{2}P̃_{2n}(x). (19)

By mathematical induction, we conclude Theorem 9. Let m be a positive odd integer, and let p = (m + 1)/2. Then

(\sqrt{x(1 − x)})^{m}

= \frac{(m!!)^{2}}{8^{p}} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}(n + 1) ⋯ (n + p − 1)(1 − 2n) ⋯ (2p − 1 − 2n)} \binom{2n}{n}^{2}P̃_{2n}(x).

In terms of the Pochhammer symbol, (a)_{k} := a(a + 1) ⋯ (a + k − 1) for k ≥ 1, the above expression can be rewritten compactly as

(\sqrt{x(1 − x)})^{m} = \frac{(m!!)^{2}}{16^{p}} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{2}P̃_{2n}(x).  (20)

Next, let f(x) = [x(1 − x)]^{−1/4}. Using Lemma 2, a_{n} = 0 for odd n. Moreover, in view of (6),

a_{2n} = (4n + 1) ∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k} ∫_{0}^{1} x^{k−1/4}(1 − x)^{−1/4} dx
= (4n + 1) ∑_{k=0}^{2n} \binom{2n}{k}\binom{2n + k}{k}(−1)^{k}B(k + 3/4, 3/4).  (21)

Using (15) and

Γ(k + 3/4) = \frac{3 · 7 ⋯ (4k − 1)}{4^{k}}Γ(3/4),

we have

B(k + 3/4, 3/4) = \frac{Γ(k + 3/4)Γ(3/4)}{Γ(k + 3/2)}

= \frac{4Γ^{2}(3/4)}{\sqrt{π}} \frac{3 · 7 ⋯ (4k − 1)(k + 1)!}{(2(k + 1))!}

= \frac{4Γ^{2}(3/4)}{\sqrt{π}} \frac{3 · 7 ⋯ (4k − 1)}{2(2k + 1)k!\binom{2n}{n}}.

Substituting this into (21) and then applying (12) yields

a_{2n} = \frac{2Γ^{2}(3/4)}{\sqrt{π}} \frac{1}{4^{n}} \binom{2n}{n}.

Notice that

Γ(3/4)Γ(1/4) = \frac{π}{sin(π/4)} = \sqrt{2} π.  (22)

Theorem 10. For x ∈ (0, 1),

\frac{1}{\sqrt[4]{x(1 − x)}} = \frac{4π^{3/2}}{Γ^{2}(1/4)} ∑_{n=0}^{∞} \frac{1}{4^{n}} \binom{2n}{n}P̃_{2n}(x).  (23)

Finally, applying (6), (13) and (22) gives

Theorem 11. For x ∈ (0, 1),

\sqrt[4]{x(1 − x)} = \frac{π^{3/2}}{2Γ^{2}(3/4)} ∑_{n=0}^{∞} \frac{4n + 1}{4^{n}(4n + 3)(1 − 4n)} \binom{2n}{n}P̃_{2n}(x).  (24)

3 Some new Ramanujan-like series

The Fourier-Legendre series expansions (14), (18), (19), (20), (23) and (24), along with their specialization, inner product and Parseval’s identity, will enable us to find many Ramanujanlike series involving powers of central binomial coefficients.

First, replacing 2x − 1 by t in (14), (16) and (18),

Theorem 12. For |t| < 1,

\frac{1}{\sqrt{1 − t^{2}}} = \frac{1}{2} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}} \binom{2n}{n}^{2} P_{2n}(t);

\frac{t}{\sqrt{1 − t^{2}}} = \frac{1}{4} ∑_{n=0}^{∞} \frac{(4n + 3)(2n + 1)π}{16^{n}(n + 1)} \binom{2n}{n}^{2} P_{2n+1}(t);

\sqrt{1 − t^{2}} = \frac{1}{8} ∑_{n=0}^{∞} \frac{(4n + 1)π}{16^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{2} P_{2n}(t).

These recover Formulas 8.922 No. 3–5 in [9, p. 988], for which I am unable to find references to their proofs. It is interesting to compare these identities with the well-known generating function of the central binomial coefficients:

\frac{1}{\sqrt{1 − t^{2}}} = ∑_{n=0}^{∞} \frac{1}{4^{n}} \binom{2n}{n}t^{2n}.

Furthermore, let x = cos^{2}(θ/2) in (14), (18)–(20), respectively. Then 2x − 1 = cos θ, and P̃_{2n}(x) = P_{2n}(cos θ).

Theorem 13. For θ ∈ (0, π),

∑_{n=0}^{∞} \frac{4n + 1}{16^{n}} \binom{2n}{n}^{2}P_{2n}(cos θ) = \frac{2}{π sin θ},  (25)

∑_{n=0}^{∞} \frac{4n + 1}{16^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{2}P_{2n}(cos θ) = \frac{8 sin θ}{π},  (26)

∑_{n=0}^{∞} \frac{4n + 1}{64^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)} \binom{2n}{n}^{2}P_{2n}(cos θ) = \frac{32 sin^{3} θ}{9π},  (27)

∑_{n=0}^{∞} \frac{4n + 1}{64^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{2}P_{2n}(cos θ) = \frac{16^{p} sin^{m} θ}{2^{m}(m!!)^{2}π} ,  (28)

where m is a positive odd integer and p = (m + 1)/2.

Identity (26) appeared in [9, p. 989] as Formula 8.925 No. 2. Next, let x = cos^{2}(θ/2) in (23) and (24), respectively.

Theorem 14. For θ ∈ (0, π),

∑_{n=0}^{∞} \frac{1}{4^{n}} \binom{2n}{n}P_{2n}(cos θ) = \frac{\sqrt{2} Γ^{2}(1/4)}{4\sqrt{sin θ}π^{3/2}} ,  (29)

∑_{n=0}^{∞} \frac{4n + 1}{4^{n}(4n + 3)(1 − 4n)} \binom{2n}{n}P_{2n}(cos θ) = \frac{\sqrt{2 sin θ} Γ^{2}(3/4)}{π^{3/2}}.  (30)

Based on Theorems 13 and 14, via specialization, we find some Ramanujan-like series as follows.

Example 1. Recall that P_{2n}(0) = (−1)^{n}\binom{2n}{n}/4^{n}. Letting θ = π/2 in Theorem 13 yields

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{64^{n}} \binom{2n}{n}^{3} = \frac{2}{π} ,  (31)

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{64^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{3} = \frac{8}{π} ,  (32)

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{64^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)} \binom{2n}{n}^{3} = \frac{32}{9π} ,  (33)

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{64^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{3} = \frac{16^{p}}{2^{m}(m!!)^{2}π} ,  (34)

where m is a positive odd integer and p = (m + 1)/2. (31) reveals Bailey’s classical formula (See [1], or [3, Formula 5.5.24, p. 184]), which was originally obtained via hypergeometric functions.

Example 2. Let θ = π/2 in Theorem 14. Then

∑_{n=0}^{∞} \frac{(−1)^{n}}{16^{n}} \binom{2n}{n}^{2} = \frac{\sqrt{2}Γ^{2}(1/4)}{4π^{3/2}},  (35)

∑_{n=0}^{∞} \frac{(−1)^{n}(4n + 1)}{16^{n}(4n + 3)(1 − 4n)} \binom{2n}{n}^{2} = \frac{\sqrt{2}Γ^{2}(3/4)}{π^{3/2}}.  (36)

We also obtain some interesting series following in another direction:

Example 3. Since P_{n}(1) = 1 for all n ≥ 0, letting θ = 0 in (30), (26)–(28), respectively, then splitting the n = 0 term, we find

∑_{n=1}^{∞} \frac{4n + 1}{4^{n}(4n + 3)(4n − 1)} \binom{2n}{n} = \frac{1}{3},  (37)

∑_{n=1}^{∞} \frac{4n + 1}{16^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{2} = −2,  (38)

∑_{n=1}^{∞} \frac{4n + 1}{16^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)} \binom{2n}{n}^{2} = −\frac{2}{3},  (39)

∑_{n=1}^{∞} \frac{4n + 1}{16^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{2} = −\frac{1}{p!(1/2)_{p}} ,  (40)

where p ∈ ℕ.

Next, let f(x) = ∑_{n=0}^{∞} a_{n}P̃_{n}(x) and g(x) = ∑_{n=0}^{∞} b_{n}P̃_{n}(x). If f(x)g(x) ∈ L^{2}[0, 1], then the inner product of f and g leads to

∑_{n=0}^{∞} (\frac{1}{\sqrt{2n + 1}} a_{n}) · (\frac{1}{\sqrt{2n + 1}} b_{n}) = ∫_{0}^{1} f(x)g(x) dx.  (41)

The inner products of (14) with (18), (14) with (19), (18) with (19) and (14) with (20) lead to the following

Theorem 15. Let m be a positive odd integer and p = (m + 1)/2. Then

∑_{n=0}^{∞} \frac{4n + 1}{256^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{4} = \frac{16}{π^{2}},  (42)

∑_{n=0}^{∞} \frac{4n + 1}{256^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)} \binom{2n}{n}^{4} = \frac{128}{27π^{2}} ,  (43)

∑_{n=0}^{∞} \frac{4n + 1}{256^{n}(n + 1)^{2}(n + 2)(1/2 − n)^{2}(3/2 − n)} \binom{2n}{n}^{4} = \frac{2048}{135π^{2}} ,  (44)

∑_{n=0}^{∞} \frac{(4n + 1)}{256^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{4} = \frac{16^{p}[(p − 1)!]^{2}}{(m!!)^{2}m!π^{2}} .  (45)

Similarly, the inner products of (14) with (23), (14) with (24), (18) with (23), (18) with (24), (19) with (23), and (19) with (24) lead to

Theorem 16.

∑_{n=0}^{∞} \frac{1}{64^{n}} \binom{2n}{n}^{3} = \frac{Γ^{4}(1/4)}{4π^{3}},  (46)

∑_{n=0}^{∞} \frac{4n + 1}{64^{n}(4n + 3)(1 − 4n)} \binom{2n}{n}^{3} = \frac{4Γ^{4}(3/4)}{π^{3}},  (47)

∑_{n=0}^{∞} \frac{1}{64^{n}(n + 1)(1/2 − n)} \binom{2n}{n}^{3} = \frac{Γ^{4}(1/4)}{3π^{3}},  (48)

∑_{n=0}^{∞} \frac{(4n + 1)}{64^{n}(n + 1)(1/2 − n)(4n + 3)(1 − 4n)} \binom{2n}{n}^{3} = \frac{48Γ^{4}(3/4)}{5π^{3}},  (49)

∑_{n=0}^{∞} \frac{1}{64^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)} \binom{2n}{n}^{3} = \frac{20Γ^{4}(1/4)}{189π^{3}},  (50)

∑_{n=0}^{∞} \frac{(4n + 1)}{64^{n}(n + 1)(n + 2)(1/2 − n)(3/2 − n)(4n + 3)(1 − 4n)} \binom{2n}{n}^{3} = \frac{448Γ^{4}(3/4)}{135π^{3}}. (51)

Furthermore, the inner product of (23) and (24) leads to a simple Ramanujan-like series:

∑_{n=0}^{∞} \frac{1}{16^{n}(4n + 3)(1 − 4n)} \binom{2n}{n}^{2} = \frac{1}{π} .  (52)

In general, taking the inner product (20) with (23) and (24), respectively, yields Theorem 17. Let m be a positive odd integer and p = (m + 1)/2. Then

∑_{n=0}^{∞} \frac{1}{64^{n}(n + 1)_{p}(1/2 − n)_{p}} \binom{2n}{n}^{3} = \frac{16^{p}(1 · 5 · (4p − 3))^{2}(2p)!Γ^{4}(1/4)}{4(m!!)^{2}(4p)!π^{3}}, (53)
∑_{n=0}^{∞} \frac{(4n + 1)}{64^{n}(n + 1)_{p}(1/2 − n)_{p}(4n + 3)(1 − 4n)} \binom{2n}{n}^{3} = \frac{16^{p+1}(3 · 7 · (4p − 1))^{2}(2p + 1)!Γ^{4}(3/4)}{2(m!!)^{2}(4p + 2)!π^{3}}. (54)

Finally, recall Parseval’s identity: If f(x) ∈ L^{2}[0, 1], then

∑_{n=0}^{∞} [\frac{1}{\sqrt{2n + 1}} a_{n}]^{2} = ∫_{0}^{1} f^{2}(x) dx.  (55)

Applying (55) to (23), (24) and (18)–(20), respectively, yields Theorem 18. Let m be a positive odd integer and p = (m + 1)/2. Then

∑_{n=0}^{∞} \frac{1}{16^{n}(4n + 1)} \binom{2n}{n}^{2} = \frac{Γ^{4}(1/4)}{16π^{2}},  (56)
∑_{n=0}^{∞} \frac{4n + 1}{16^{n}(4n + 3)^{2}(1 − 4n)^{2}} \binom{2n}{n}^{2} = \frac{Γ^{4}(3/4)}{2π^{2}},  (57)
∑_{n=0}^{∞} \frac{4n + 1}{256^{n}(n + 1)^{2}(1/2 − n)^{2}} \binom{2n}{n}^{4} = \frac{128}{3π^{2}},  (58)
∑_{n=0}^{∞} \frac{4n + 1}{256^{n}(n + 1)^{2}(n + 2)^{2}(1/2 − n)^{2}(3/2 − n)^{2}} \binom{2n}{n}^{4} = \frac{16384}{2835π^{2}} ,  (59)
∑_{n=0}^{∞} \frac{(4n + 1)}{256^{n}(n + 1)_{p}^{2}(1/2 − n)_{p}^{2}} \binom{2n}{n}^{4} = \frac{256^{p}(m!)^{2}}{(m!!)^{4}(2m + 1)!π^{2}} . (60)

4 Remarks

We now conclude this paper with two remarks.

1. Recall the complete elliptical integral of the first kind defined by

K(k) := ∫_{0}^{π/2} \frac{dθ}{\sqrt{1 − k^{2} sin^{2} θ}}.

Invoking the generating function of the central binomial coefficients

∑_{n=0}^{∞} \frac{1}{4^{n}} \binom{2n}{n}x^{2n} = \frac{1}{\sqrt{1 − x^{2}}} for |x| < 1

and Wallis’ formula

∫_{0}^{π/2} sin^{2n} θ dθ = \frac{π}{2^{2n+1}} \binom{2n}{n},

we have

K(k) = \frac{π}{2} ∑_{n=0}^{∞} \frac{1}{16^{n}} \binom{2n}{n}^{2}k^{2n}.

Moreover, Watson’s elliptic function identity [3, p. 188] claims

\frac{π^{2}}{4} ∑_{n=0}^{∞} \frac{1}{64^{n}} \binom{2n}{n}^{3} = K^{2}(1/\sqrt{2}).

Thus, together with (46), we obtain another proof of the classical result [3, Theorem 1.7, p. 25]:

K (\frac{1}{\sqrt{2}}) = \frac{Γ^{2}(1/4)}{4\sqrt{π}}.

2. The neat forms of Fourier-Legendre series in Section 3 rely on the combinatorial identities (9), (12) and (13). Similar to (9), by the WZ-method, Tauraso [15, Lemma 4.2] proved

∑_{k=0}^{2n} (−1)^{k}\binom{2n}{k}\binom{2n + k}{k}\binom{2k}{k}4^{2n−k}H_{k} = \binom{2n}{n}^{2}H_{2n}.

This enables us to find Ramanujan-like series and other well-known constants involving the product of the central binomial coefficients and the harmonic numbers like

∑_{n=1}^{∞} \frac{1}{16^{n}(2n − 1)}\binom{2n}{n}^{2}H_{2n} = \frac{6 ln 2 − 2}{π}

and

∑_{n=0}^{∞} \frac{1}{16^{n}(2n + 1)}\binom{2n}{n} (3H_{2n+1} + \frac{4}{2n + 1}) = 8G,

where G is the Catalan constant defined by ∑_{n=0}^{∞}(−1)^{n}/(2n + 1)^{2}. The interested reader is encouraged to pursue results in this direction.

Addendum. After this paper was submitted for publication, the referee brought my attention to the papers [12] and [5]. The identities (33), (34), and (58)–(60) are found in [12], in which Levrie used a different approach to derive the Fourier-Legendre series (20). The identities (58)–(60) were also rediscovered by Cantarini and D’Aurizio [5].

5 Acknowledgments

Thanks to the anonymous referee for useful comments and bringing to my attention the papers [12] and [5]. Also thanks to Dr. Bradie and the editor-in-chief for their suggestions which have made this paper more reader-friendly.

References

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[14] R. Tauraso, Personal communication, 2021.

[15] R. Tauraso, Supercongruences related to _{3}F_{2}(1) involving harmonic numbers, Int. J. Number Theory 14 (2018), 1093–1109.

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2020 Mathematics Subject Classification: Primary 11B83; Secondary 05A10. Keywords: central binomial coefficient, Legendre polynomial, Ramanujan-like series, Fourier-Legendre series.

(Concerned with sequences \underline{A000045}, \underline{A000984}, \underline{A001008}, and \underline{A005408}.)

Received October 8 2021; revised version received January 5 2022. Published in Journal of Integer Sequences, January 28 2022.

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