Bauer's series for $\frac{1}{\pi}$

Question

Recently, someone asked a question involving the expression $$ \sum_{n=0}^{\infty} (-1)^n (4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3 $$

At first glance, I knew that the expression was the value of a hypergeometric function, and anticipated difficulty in determining its exact value. Surprisingly, the value obtained using Mathematica was $\frac{2}{\pi}$. He couldn't recall the context in which the formula appeared well. (He mentioned seeing the expression in a comic book!) However, at this point, I suspect this may be a series studied by Ramanujan. Interestingly, when variables are introduced into the expression, as the following $$ \sum_{n=0}^{\infty} x^n (4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3 $$ a very complicated hypergeometric function emerges, Nonetheless, Mathematica was unable to simplify the expression obtained by substituting $x=-1$ to revert to the original form.

Therefore, I believe that the result might be derived from something like a complex contour integral or a similar method, and might not be readily analyzed as a special value of a hypergeometric function.

Please help in understanding the reasoning behind the evaluation of this series.

Answer 1

This series is of a different breed in comparison with others you can find in the famous paper Modular equations and approximations to $\pi$ where Ramanujan used elliptic functions to calculate similar series.

Instead, your series appears in the chapter one of Hardy's book: Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work with the number (1.2)

$$ 1-5\left(\frac{1}{2}\right)^3 + 9\left(\frac{1\cdot 3}{2\cdot 4}\right)^3-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^3+... = \frac{2}{\pi} \tag{1.2} $$

In that book Hardy gave a hint of how the formula can be proven:

The series formulae (1.1)-(1.4) I found much more intriguing, and it soon became obvious that Ramanujan must possess much more general theorems and was keeping a great deal up his sleeve. The second (1.2) is a formula of Bauer well known in the theory of Legrende series, but the others are much harder than they look.

Following Hardy's advice, the relation can be proved using the theory of Legendre polynomials.

Consider the Legrende polynomials defined with the Rodrigues’ formula:

$$ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n \tag{1} $$

If $f(x)$ is a very smooth function in $|x|\leq 1$ then it can be expanded in terms of Legendre polynomials:

$$ f(x) = \sum_{n=0}^{\infty} a_nP_n(x) $$

where

$$ a_n = \frac{2n+1}{2} \int_{-1}^{1} P_n(x) f(x) dx $$

Consider function $\displaystyle f(x) = \frac{1}{\sqrt {1-x^2}}$. We will show that this function has the following expansion in Legendre polynomials:

$$ \frac{1}{\sqrt{1-x^2}} = \frac{\pi}{2} \left(1 + 5\left(\frac{1}{2}\right)^2 P_{2}(x) +9\left(\frac{1\cdot 3}{2\cdot 4}\right)^2P_4(x) + 13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2 P_6(x)+... \right) $$

First, we have to calculate the coefficients:

$$a_n = \frac{2n+1}{2} \int_{-1}^{1} \frac{P_n(x)}{\sqrt{1-x^2}}dx \tag{*}$$

This can be proven by induction:

$$ \int_{-1}^1 \frac{P_n(t)dt}{\sqrt{1-t^2} } = \begin{cases} \displaystyle \pi \left[\frac{(n-1)!!}{n!!}\right]^2 & \quad n=2m , \;\; m\in \mathbb{Z}_{\geq 0}\\ \displaystyle 0 &\quad n=2m+1, \;\; m\in \mathbb{Z}_{\geq0} \end{cases}$$

For that end, these properties of the Legrende polynomials are needed:

$$ P_n(x) = \frac{2n-1}{n} xP_{n-1}(x) - \frac{n-1}{n} P_{n-2}(x) \tag{2}$$

$$ \frac{d}{dx} P_n(x) = \frac{n}{1-x^2} \left[P_{n-1}(x) -xP_n(x) \right] \tag{3} $$

Base case. $m=0$

By the Rodrigues’ formula:

\begin{eqnarray*} \int_{-1}^1 \frac{P_0(t)dt}{\sqrt{1-t^2} } &=& \int_{-1}^1 \frac{dt}{\sqrt{1-t^2} }\\ & =& \arcsin(1)-\arcsin(-1) = \pi \end{eqnarray*}

\begin{eqnarray*} \int_{-1}^1 \frac{P_1(t)dt}{\sqrt{1-t^2} } &=& \int_{-1}^1 \frac{tdt}{\sqrt{1-t^2} }\\ & =& \left[-\sqrt{1-x^2}\right]_{-1}^1 = 0 \end{eqnarray*}

Inductive step. Suppose this is valid for $m=k$

$$ \int_{-1}^1 \frac{P_n(t)dt}{\sqrt{1-t^2} } = \begin{cases} \displaystyle \pi \left[\frac{(n-1)!!}{n!!}\right]^2 & \quad n=2k \\ \displaystyle 0 &\quad n=2k+1 \end{cases}$$

We have to prove it for $m=k+1$

From the recurrence formula (2) and the inductive step:

\begin{eqnarray*} \int_{-1}^1 \frac{P_{2k+2}(t)dt}{\sqrt{1-t^2} } &=& \frac{4k+3}{2k+2}\int_{-1}^1 t\frac{P_{2k+1}(t)dt}{\sqrt{1-t^2} }- \frac{2k+1}{2k+2} \int_{-1}^1 \frac{P_{2k}(t)dt}{\sqrt{1-t^2} } \\ &=& \frac{4k+3}{2k+2}\int_{-1}^1 \frac{tP_{2k+1}(t)dt}{\sqrt{1-t^2} }- \pi \frac{(2k+1)!!}{(2k+2)!!}\frac{(2k-1)!!}{(2k)!!} \\ \end{eqnarray*}

Now integrating by parts and using (3):

\begin{eqnarray*} \int_{-1}^1 \frac{tP_{2k+1}(t)dt}{\sqrt{1-t^2} } &=& (2k+1)\int_{-1}^1 \sqrt{1-t^2}P'_{2k+1}(t)dt \\ &=& (2k+1)\int_{-1}^1 \frac{P_{ 2k}(t)}{\sqrt{1-t^2}}dt - (2k+1)\int_{-1}^1 \frac{tP_{ 2k+1}(t)}{\sqrt{1-t^2}}dt \end{eqnarray*}

Hence, by the inductive step:

\begin{eqnarray*} \int_{-1}^1 \frac{tP_{2k+1}(t)dt}{\sqrt{1-t^2} } &=& \frac{ (2k+1)}{(2k+2)} \int_{-1}^1 \frac{P_{ 2k}(t)}{\sqrt{1-t^2}}dt = \pi \frac{(2k+1)!!}{(2k+2)!!} \frac{(2k-1)!!}{(2k)!!} \end{eqnarray*}

Therefore:

\begin{eqnarray*} \int_{-1}^1 \frac{P_{2k+2}(t)dt}{\sqrt{1-t^2} } &=& \frac{4k+3}{2k+2}\int_{-1}^1 t\frac{P_{2k+1}(t)dt}{\sqrt{1-t^2} }- \frac{2k+1}{2k+2} \int_{-1}^1 \frac{P_{2k}(t)dt}{\sqrt{1-t^2} } \\ &=& \pi \frac{(4k+3)}{(2k+2)} \frac{(2k+1)!!}{(2k+2)!!} \frac{(2k-1)!!}{(2k)!!} - \pi \frac{(2k+1)!!}{(2k+2)!!}\frac{(2k-1)!!}{(2k)!!} \\ &=& \pi \left[ \frac{(2k+1)!!}{(2k+2)!!}\right]^2 \end{eqnarray*}

In a similar way you can show as an exercise that

\begin{eqnarray*} \int_{-1}^1 \frac{P_{2k+3}(t)dt}{\sqrt{1-t^2} } &=& 0 \end{eqnarray*}

Hence, from (*) we have:

$$ a_{2m} = \frac{(4m-1)\pi}{2} \left[\frac{(2m-1)!!}{(2m)!!}\right]^2 $$

$$ a_{2m+1} =0$$ Then, we have proved:

$$ \frac{1}{\sqrt{1-x^2}} = \frac{\pi}{2} \left(1 + 5\left(\frac{1}{2}\right)^2 P_{2}(x) +9\left(\frac{1\cdot 3}{2\cdot 4}\right)^2P_4(x) + 13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2 P_6(x)+... \right) $$

If $x=0$.

$$ \frac{2}{\pi} =1 + 5\left(\frac{1}{2}\right)^2 P_{2}(0) +9\left(\frac{1\cdot 3}{2\cdot 4}\right)^2P_4(0) + 13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2 P_6(0)+... $$

by the recurrence formula (2) with $x=0$ :

$$ P_{2m}(0) = -\frac{2m-1}{2m} P_{2m-2}(0) $$

Iterating

$$ P_{2m}(0) = (-1)^{m}\frac{(2m-1)}{2m}\frac{(2m-3)}{(2m-2)}\cdots \frac{1}{2} = (-1)^{m} \frac{(2m-1)!!}{(2m)!!}$$

Finally

$$ \frac{2}{\pi}= 1-5\left(\frac{1}{2}\right)^3 + 9\left(\frac{1\cdot 3}{2\cdot 4}\right)^3-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^3+... $$

Answer 2

We give an alternative approach for the series you are asking. It's known as Bauer's series and was sent by Ramanujan to Hardy in one of his letters. The beatiful proof offered by @Bertrand87 it is how Bauer obtained it.

Then Ramanujan obtained it using Clausen's formula and elliptic functions and modular equations.

There are four type of proof as I know.

1. Fourier-Legendre expansions.
2. Modular equations and elliptic functions.
3. WZ-Method.
4. Elliptic Integrals in a classical approach.

I will share the classical approach based on series and elliptic integrals. Let be: $$K(k)=\int_{0}^{1}\frac{dx}{\sqrt{1-k^2\sin^2x}},\hspace{1cm}E(k)=\int_{0}^{1}\sqrt{1-k^2\sin^2x}dx\tag{1}$$ the complete elliptic integrals of the first and second kind and $k'=\sqrt{1-k^2}$ the complementary modulus. This series is related to this fact: $$\frac{K(s')}{K(s)}=\sqrt{2}\tag{2} \implies s=\sqrt{2}-1$$ and: $$8E(s)K(s)-4\sqrt{2}K^2(s)=\sqrt{2}\pi\tag{3}.$$ A proof can be found here: Specific values of Elliptic integral of the second kind $E$.

The next we do is invoke Clausen's formula in this version (there a lot of versions of this formula, the most known is this https://en.wikipedia.org/wiki/Clausen%27s_formula): \begin{align} K^2({k})=\frac{\pi^2}{4(1-k^2)}\sum_{n=0}^{\infty}\frac{(-1)^nk^{2n}(k+1)^{-2n}(4(k-1))^{-2n}(2n)!^3}{n!^6}.\tag{4} \end{align}

And from $(4)$, derivation produces: \begin{align} K(k)E(k)=\frac{\pi^2}{4(1-k^2)}\sum_{n=0}^{\infty}\frac{(-1)^nk^{2n}(k+1)^{-2n}(4(k-1))^{-2n}(nk^2+n+1)(2n)!^3}{n!^6},\tag{5} \end{align} where we have used the relation $\frac{\mathrm{dK}}{\mathrm{d}k}=\frac{E(k)}{k(1-k^2)} - \frac{K(k)}{k}$. Then rewritting $(3)$ in terms of series using $(4)$ and $(5)$ when $k=s=\sqrt{2}-1$ we have: $$4\sqrt{2}K^2(s)=\pi^2\left(\frac{\sqrt{2}}{2}+1\right)\sum_{n=0}^{\infty}\frac{(2n)!^3(-1)^n}{2^{6n}n!^6},$$ and $$8E(s)K(s)=\pi^2\left(\sqrt{2}+1\right)\sum_{n=0}^{\infty}\frac{(n(4-2\sqrt{2})+1)(2n)!^3(-1)^n}{2^{6n}n!^6}.$$ Finally we obtain Bauer's series: $$\sqrt{2}\pi=8E(s)K(s)-4\sqrt{2}K^2(s)=\frac{\sqrt{2}\pi^2}{2}\sum_{n=0}^{\infty}\frac{(4n+1)(2n)!^3(-1)^n}{2^{6n}n!^6} \\ \implies \sum_{n=0}^{\infty}\frac{(4n+1)(2n)!^3(-1)^n}{2^{6n}n!^6}=\frac{2}{\pi}.$$

Note: $$n!!=\frac{(2n)!}{2^{n}n!}.$$

Answer 3

$$S=\sum_{n=0}^{\infty} (-1)^n\, (4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3$$ $$S=\frac{1}{\pi ^{3/2}}\sum_{n=0}^{\infty}(-1)^n\, (4n+1)\left(\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\right)^3 $$ So, an hypergeometric function is expected. $$S=\left(\frac{\Gamma \left(\frac{9}{8}\right)}{\Gamma \left(\frac{7}{8}\right) \Gamma \left(\frac{5}{4}\right)}\right)^2-\frac 12 \,\, _3F_2\left(\frac{3}{2},\frac{3}{2},\frac {3}{2};2,2;-1\right)=\frac 2 \pi$$ Now $$T=\sum_{n=0}^{\infty} (-1)^n\, (4n+1) \left(\frac{(2n-1)!!}{(2n)!!}\right)^3\, x^n$$ $$T=\frac 4 {\pi^2}\Bigg(K\left(\frac{1}{2} \left(1-\sqrt{x+1}\right)\right)\Bigg)^2-\frac x2 \,\,, _3F_2\left(\frac{3}{2},\frac{3}{2},\frac {3}{2};2,2;-x\right)$$ Checked numerically $$\frac 2 \pi\,K\left(\frac{1}{2} \left(1-\sqrt{2}\right)\right)=\frac{\Gamma \left(\frac{9}{8}\right)}{\Gamma \left(\frac{7}{8}\right) \Gamma \left(\frac{5}{4}\right)}$$

All of the above looks very consistent.