Conjecture: $\frac1\pi=\sum_{n=0}^\infty\left((n+1)\frac{C_n^3}{2^{6n}}\sum_{k=0}^n(-1)^k{n\choose k}{\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\right)$

Question

Let $C_n$ denote the $n$-th Catalan number defined by

$${\displaystyle C_{n}={\frac {1}{n+1}}{2n \choose n}=\prod \limits _{k=2}^{n}{\frac {n+k}{k}}\quad \left(n\geqslant 0\right).}$$ Next, we define the sequence $${\displaystyle A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}}.$$ I've numerically managed to verify that \begin{equation}\tag{1}\label{pi} {\displaystyle \sum _{n=0}^{\infty }(n+1)A_{n}={\frac {1}{\pi}}}. \end{equation} Is it possible to prove the relation in (\ref{pi})? If yes, then how could we go about it?

Also, is this series already known or studied in the literature? If yes, then any references will be highly appreciated. Thanks!

Answer 1

$$\sum _{k=0}^{n}(-1)^{k}{n \choose k}{\frac {(n-k)(k-1)}{(2k-1)(2k+1)}}=\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}$$ $$ A_{n}={\frac {C_{n}^{3}}{2^{6n}}}\frac{\sqrt{\pi } \,\,\Gamma (n+2)}{4 \,\Gamma \left(n+\frac{1}{2}\right)}=\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi \Gamma (n+2)^2}$$ $$(n+1)A_n=\frac{(n+1) \Gamma \left(n+\frac{1}{2}\right)^2}{4 \pi \Gamma (n+2)^2}$$ $$S_p=\sum_{n=0}^p(n+1)A_n=\frac 1 \pi \frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}$$ Now, Stirling approximation $$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=\log(p+1)+2\log \left(\Gamma \left(p+\frac{3}{2}\right)\right)-2\log \left(\Gamma \left(p+{2}\right)\right)$$ $$\log \left(\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}\right)=-\frac{1}{4 p}+\frac{1}{4 p^2}+O\left(\frac{1}{p^3}\right)$$

$$\frac{(p+1) \Gamma \left(p+\frac{3}{2}\right)^2}{\Gamma (p+2)^2}=1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right)$$

$$S_p=\frac 1 \pi \left(1-\frac{1}{4 p}+\frac{9}{32 p^2}+O\left(\frac{1}{p^3}\right) \right)$$

Answer 2

We show OP's identity has a representation \begin{align*} \color{blue}{\sum_{n=0}^\infty\frac{1}{4^{2n+1}}\,\frac{1}{n+1}\binom{2n}{n}^2=\frac{1}{\pi}}\tag{1} \end{align*} OP is asking for references and actually this series is well known and has an interesting history. It was Ramanujan who sent a generalisation of (1) to Hardy. This generalised form can be written, given $q$ a positive integer as \begin{align*} \color{blue}{S(q)=\sum_{n=0}^\infty\frac{1}{4^{2n}}\,\frac{1}{n+q}\binom{2n}{n}^{2}=\frac{4^{2q}}{\pi q^2\binom{2q}{q}^2}\sum_{j=0}^{q-1}\frac{\binom{2j}{j}^2}{4^{2j}}} \end{align*} Note, in (1) we consider the special case \begin{align*} S(1)=\sum_{n=0}^\infty\frac{1}{4^{2n}}\,\frac{1}{n+1}\binom{2n}{n}^{2}=\frac{4}{\pi} \end{align*} A generalisation of $S(q)$ with a nice reference to the history of $S(q)$ can be found in A Certain Series Associated with Catalan’s Constant by V. S. Adamchik.

In order to show (1) we simplify OP's $A_n$ with elementary means. The following is valid for $n\geq 0$: \begin{align*} \color{blue}{\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}=(n+1)4^{n-1}\binom{2n}{n}^{-1}}\tag{2} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)\left(\frac{3}{4}\,\frac{1}{2k+1}-\frac{1}{4}\,\frac{1}{2k-1}\right)\\ &=n\sum_{k=0}^n(-1)^k\binom{n-1}{k}\left(\frac{3}{4}\,\frac{1}{2k+1}-\frac{1}{4}\,\frac{1}{2k-1}\right)\\ &=\frac{3n}{4}\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\frac{1}{2k+1}-\frac{n}{4}\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\frac{1}{2k-1}\tag{3} \end{align*}

Now we calculate the left-hand sum of (3) following an approach from problem section, chapter 1 in Combinatorial Identities by J. Riordan. We derive a recurrence relation by calculating $f_n$ in two ways.

\begin{align*} f_n&=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2^{k+1}}\\ &=1+\sum_{k=1}^n(-1)^k\binom{n-1}{k}\frac{1}{2k+1}+\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\frac{1}{2k+1}\\ &=f_{n-1}+\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\frac{1}{2k+1}\tag{4.1}\\ \\ f_n&=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(1-\frac{2k}{2k+1}\right)\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}-2\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{k}{2k+1}\\ &=\delta_{n,0}-2n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\frac{1}{2k+1}\tag{4.2} \end{align*}

We combine (4.1) and (4.2) and obtain for $n\geq 1$ \begin{align*} f_0&=1\\ f_n&=\delta_{n,0}-2n\left(f_n-f_{n-1}\right)\\ (2n+1)f_n&=2nf_{n-1}+\delta_{n,0}\\ f_n&=\frac{2n}{2n+1}f_{n-1}=\cdots=\frac{(2n)!!}{(2n+1)!!}f_0=\frac{(2n)!!(2n)!!}{(2n+1)!}\\ &=\frac{2^nn!2^nn!}{(2n+1)!}=\frac{4^n}{2n+1}\binom{2n}{n}^{-1} \end{align*}

It follows \begin{align*} \color{blue}{f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} =\frac{4^n}{2n+1}\binom{2n}{n}^{-1}} \end{align*}

In the same way we can calculate the right-hand sum of (3.3) and obtain \begin{align*} \color{blue}{g_n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{2k-1}=-4^n\binom{2n}{n}^{-1}} \end{align*}

Putting $f_n$ and $g_n$ in (3) we obtain after some simplifications \begin{align*} \color{blue}{\sum_{k=0}^n(-1)^k}&\color{blue}{\binom{n}{k}\frac{(n-k)(k-1)}{(2k-1)(2k+1)}}\\ &=\frac{3n}{4}f_{n-1}-\frac{n}{4}g_{n-1}\\ &=\frac{3n}{4}\,\frac{4^{n-1}}{2n-1}\binom{2n-2}{n-1}^{-1}+\frac{n}{4}\,4^{n-1}\binom{2n-2}{n-1}^{-1}\\ &\,\,\color{blue}{=(n+1)4^{n-1}\binom{2n}{n}^{-1}=4^{n-1}C_{n}^{-1}}\tag{5} \end{align*} and the claim (2) follows.

With (5) we calculate $A_n$ as \begin{align*} \color{blue}{A_n}&=\frac{C_n^3}{2^{6n}}4^{n-1}(n+1)C_n^{-1}=\frac{C_n^2}{4^{2n+1}}(n+1)\\ &\,\,\color{blue}{=\frac{1}{4^{2n+1}}\,\frac{1}{(n+1)}\binom{2n}{n}^2} \end{align*}

so that OPs identity can be stated as \begin{align*} \color{blue}{\sum_{n=0}^\infty\frac{1}{4^{2n+1}}\,\frac{1}{n+1}\binom{2n}{n}^2=\frac{1}{\pi}} \end{align*} and the claim (1) follows.