Prove: $\lim_{n\to\infty}{\sum_{m=0}^{n}{\sum_{k=0}^{n-m}{\frac{2^{n-m-k}}{n-m+1}\,\frac{{{2k}\choose{k}}{{2m}\choose{m}}}{{{2n}\choose{n}}}}}}=\pi$

Solution 1:

I think OP did a really good job and this answer aims to indicate that it is plausible to obtain the specific type of generating functions like $\arctan$ as stated by OP. Here we start with the binomial expression \begin{align*} q_n:=\sum_{m=0}^n\sum_{k=0}^{n-m}\binom{2k}{k}\binom{2m}{m}\frac{2^{n-m-k}}{n-m+1}\tag{1} \end{align*} which corrresponds to OPs limit expression without the factor $\binom{2n}{n}^{-1}$ and derive from it a generating function.

Note that since $q_n=\frac{1}{4^n}r_{n+1}$ OPs claim can be stated as \begin{align*} q_n\sim \pi\binom{2n}{n}\sim\sqrt{\frac{\pi}{n}}\cdot 4^n \end{align*} where we use the asymptotic formula of the central binomial coefficient.

Two aspects:

• We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. Recalling the generating function of the central binomial coefficient we can write for instance \begin{align*} [z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{2} \end{align*}

• We can sum up coefficients $a_n$ by multiplication with $\frac{1}{1-z}$. If $A(z)=\sum_{n=0}^\infty a_nz^n$ we have \begin{align*} \frac{1}{1-z}A(z)&=\sum_{n=0}^\infty\left( \sum_{k=0}^na_k\right)z^n \end{align*} Somewhat more general by multiplication with $\frac{1}{1-pz}$ we have \begin{align*} \frac{1}{1-pz}A(z)&=\sum_{n=0}^\infty \left(\sum_{k=0}^na_kp^{n-k}\right) z^n\tag{3} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{m=0}^n}&\color{blue}{\sum_{k=0}^{n-m}\binom{2k}{k}\binom{2m}{m}\frac{2^{n-m-k}}{n-m+1}}\\ &=\int_{0}^1\sum_{m=0}^n\sum_{k=0}^{n-m}\binom{2k}{k}\binom{2m}{m}2^{n-m-k}z^{n-m}\,dz\tag{4}\\ &=\int_{0}^1\sum_{m=0}^n\binom{2m}{m}z^{n-m}\sum_{k=0}^{n-m}\binom{2k}{k}2^{n-m-k}\,dz\tag{5}\\ &=\int_{0}^1\sum_{m=0}^n\binom{2m}{m}z^{n-m}[t^{n-m}]\frac{1}{(1-2t)\sqrt{1-4t}}\,dz\tag{6}\\ &=\int_{0}^1\sum_{m=0}^\infty\binom{2n-2m}{n-m}z^m[t^m]\frac{1}{(1-2t)\sqrt{1-4t}}\,dz\tag{7}\\ &=\int_{0}^\infty\sum_{m=0}^\infty[u^{n-m}]\frac{1}{\sqrt{1-4u}}z^m[t^m]\frac{1}{(1-2t)\sqrt{1-4t}}\,dz\tag{8}\\ &=[u^n]\frac{1}{\sqrt{1-4u}}\int_{0}^1\sum_{m=0}^\infty(zu)^m[t^m]\frac{1}{(1-2t)\sqrt{1-4t}}\,dz\tag{9}\\ &=[u^n]\frac{1}{\sqrt{1-4u}}\int_{0}^1\frac{1}{(1-2zu)\sqrt{1-4zu}}\,dz\tag{10}\\ &=[u^n]\frac{1}{\sqrt{1-4u}}\left.\frac{\arctan\left(\sqrt{1-4zu}\right)}{u}\right|_{z=0}^{z=1}\tag{11}\\ &\,\,\color{blue}{=[u^{n+1}]\frac{1}{\sqrt{1-4u}}\left(-\arctan\left(\sqrt{1-4u}\right)+\frac{\pi}{4}\right)}\tag{12} \end{align*} and we also get when deriving generating functions directly from (1) the same function as OP. The scaling factor $4$ in $\sqrt{1-4u}$ is from formula (2) and indicates the connection between $q_n$ and $r_n$ as stated at the beginning of this post.

Comment:

• In (4) we use $\frac{1}{p+1}=\int_0^1z^{p}\,dz$ where $p\ne -1$.

• In (5) we do a rearrangement only.

• In (6) we apply the coefficient of operator by using (2) and (3) with $p=2$.

• In (7) we change the order of summation by $m\to n-m$ and we replace the upper index $n$ by $\infty$ without changing anything, since $\binom{2n-2m}{n-m}=0$ when $m>n$.

• In (8) we apply again the coefficient of operator to $\binom{2n-2m}{n-m}$ according to (2).

• In (9) we use the linearity of the operators and apply the rule $[u^{p-q}]A(u)=[u^p]u^qA(u)$.

• In (10) we apply the substitution rule of the coefficient of operator with $t=zu$
  \begin{align*} A(z)=\sum_{m=0}^\infty a_m z^m=\sum_{m=0}^\infty z^m [u^m]A(u) \end{align*}

• In (11) we integrate obtaining the $\arctan$ function.

• In (12) we finally evaluate the $\arctan$ function at lower and upper limit and apply again the rule $[u^n]\frac{1}{u}A(u)=[u^{n+1}]A(u)$ as we did in (10).