How to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}$

Question

I want to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}.$ I tried from arcsin(x) series and got $\frac{1-z^4}{(1+z^4)^{\frac{2}{3}}}= 1-5(\frac{1}{2})z^4+9(\frac{(1)(3)}{(2)(4)})z^8-13(\frac{(1)(3)(5)}{(2)(4)(6)})z^{12}+...$ ,but i got stuck with $\frac{(2n-1)!!}{2n!!}$ terms and I don't know how to get $(\frac{(2n-1)!!}{2n!!})^3 $ Can anyone help me to Evaluate this infinite sum? You can share your own way.

Answer 1

$$\frac{2}{\pi}=\sum_{n=0}^{\infty}(-1)^n\binom{2n}{n}^3\left(\frac{1+4n}{2^{6n}}\right)$$

One can prove this using Fourier-Legendre Expansions thereby completely bypassing the Elliptic Integral Pathway.

First denote,

$$P_s:=P_s(k)={_2F_1}\left[\begin{array}c-s,1+s\\1\end{array}\middle|\,k^2\right]$$

and $P'_s:=P_s(\sqrt{1-k^2})$

which represents the Alternate Elliptic Integrals for $$-s\in\left\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{6}\right\}$$

But that won't be the focus here, instead we will be using this when $s$ is an integer.

Using Differential Equations, one may prove that

$$\int_0^1 kP_a'P_b\ dk=\frac{\sin \pi b-\sin \pi a}{2\pi[b(b+1)-a(a+1)]}$$

and we can also prove that if $n$ is an integer then $P_n'=(-1)^nP_n$

And we also have the evaluation of $P_s$ at $k=1/\sqrt{2}$

$$P_s\left(\frac{1}{\sqrt{2}}\right)=\frac{\cos(\pi s/2)}{2^{s}}\binom{s}{s/2}$$

This results in for $a$ and $b$ integer,

$$\int_0^1kP_aP_b\ dk=0,\quad a\neq b$$

$$\int_0^1kP^2_n\ dk=\frac{1}{2(2n+1)}$$

which is basically an orthogonality property.

So one may write for a function $f(k)$

$$f(k) \sim2\sum_{n=0}^{\infty}(2n+1)\left[\int_0^1kf(k)P_n\ dk\right]P_n $$

Those familiar with Fourier Legendre Expansions will realize it's the same thing.

Using Clausen's Formula we can prove that,

$$\int_0^{\pi/2}P_s(2kk'\sin t)\ dt=\frac{\pi P^2_s}{2}$$

or evaluating it at $k=1/\sqrt{2}$ gives

$$\int_0^{\pi/2}P_s(\sin t)\ dt=\frac{\pi}{2}\frac{\cos(\pi s/2)^2}{2^{2s}}\binom{s}{s/2}^2$$

but this integral can be rewritten as,

$$\int_0^1k\left[\frac{1}{k\sqrt{1-k^2}}\right]P_{2n}\ dk=\frac{\pi}{2}\frac{1}{2^{4n}}\binom{2n}{n}^2$$

which gives us the Fourier Legendre Expansion,

$$\frac{1}{k\sqrt{1-k^2}}\sim\pi\sum_{n=0}^{\infty}\binom{2n}{n}^2\left(\frac{1+4n}{2^{4n}}\right)P_{2n}$$

This doesn't converge so I am not sure of this last step but if we put $k=1/\sqrt{2}$ and use

$$P_{2n}\left(\frac{1}{\sqrt{2}}\right)=\frac{(-1)^n}{2^{2n}}\binom{2n}{n}$$

we will recover back the series for $1/\pi$.

I have come up with this proof myself no I am not sure of its validity.

I hope someone can verify.

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EDIT

It seems that if the following series converges absolutely

$$\sum_{n=0}^{\infty}c_n$$

then the series of functions $$\sum_{n=0}^{\infty}c_nP_n$$ converges pointwise.

So in our case as $$\sum_{n=0}^{\infty}\binom{2n}{n}^2\left(\frac{1+4n}{2^{4n}}\right)$$

does not converge then the series of functions does not converge to $\frac{1}{k\sqrt{1-k^2}}$.

But it is still interesting so I will leave this answer here as reference for future readers.

Answer 2

We prove in this answer Bauer's series that was known since 1859. The original proof uses Fourier-Legendre expansions as I know. This series appears in one of the letters that Ramanujan sent to Hardy. We want to show an elementary proof of it.

We begin with Legendre's relation with the constraint: \begin{align} \frac{K(k')}{K(k)}=\sqrt{r}\tag{1}, \end{align} where $k'=\sqrt{1-k^2}$ is the complementary modulus. Then we can rewrite Legendre's relation in this way: \begin{align} \frac{\pi}{2\sqrt{r}}=\frac{K(k')E(k')}{r}+E(k)K(k)-K^2({k})\tag{2}. \end{align} I we put $r=2$ in $(1)$ then it is well known that $k=\sqrt{2}-1$. So replacing in $(2)$ we have: \begin{align} \frac{\pi}{2\sqrt{2}}=\frac{K(k')E(k')}{2}+E(k)K(k)-K^2(k).\tag{3} \end{align} Now we are going to invoke the next series representions for $K^2(k)$, $K(k)E(k)$ and $K(k')E(k')$ namely:

\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} \begin{align} K(k)E(k)=\frac{\pi^2}{4(1-k^2)}\sum_{n=0}^{\infty}\frac{(-1)^{n}k^{2n}(4(k-1))^{-2n}(k+1)^{-2n}(nk^2+n+1)(2n)!^3}{n!^6},\tag{5} \end{align} and \begin{align} K(k')E(k')=\frac{\pi^2(k^2+1)}{8}\sum_{n=0}^{\infty}\frac{(-1)^{n}k^{-2n-1}(1-k^2)^{2n}(2n+1)(2n)!^3}{2^{8n}n!^6}.\tag{6} \end{align} We have to note that $(4)$ and $(5)$ are valid if $k \in (0,\frac{1}{\sqrt{2}})$ and $(6)$ is valid only if $k \in [\sqrt{2}-1,\sqrt{2}+1]$. Under this conditions using $(3)$, $(4)$, $(5)$ and $(6)$ when $k=\sqrt{2}-1$ then: \begin{align} \frac{\pi}{2\sqrt{2}}=\frac{\sqrt{2}\pi^2}{8}\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}. \end{align} Which is Bauer's series.

Remarks

To obtain $(4)$,$(5)$ and $(6)$ we begin with the formula $(1)$ given in this answer: (Integrals of elliptic integrals.)

• $(4)$ is obtained from $(9)$ in the linked answer using Landen's transformation.
• $(5)$ is obtained from $(1)$ in the linked answer using Landen's transformation.
• $(6)$ is obtained replacing $k=\sqrt{1-k^2}$ in formula $(1)$ of the linked answer.