Consider the following limit: $$\lim_{n\to\infty}{\sum_{m=0}^{n}{\sum_{k=0}^{n-m}{\left[\frac{2^{n-m-k}}{n-m+1}\,\frac{{{2k}\choose{k}}{{2m}\choose{m}}}{{{2n}\choose{n}}}\right]}}}=\pi$$ I cooked this up while playing around with power series (details below).

Is there a more direct way to prove this limit?

Consider the functions $f(x)=\frac{\tan^{-1}(\sqrt{1-x})}{\sqrt{1-x}}$, $g(x)=\frac{\pi/2-\tan^{-1}(\sqrt{1-x})}{\sqrt{1-x}}$. We write the power series: $$f(x)=\sum_{n=0}^{\infty}{(s_n\pi-r_n)x^n},\qquad{g}(x)=\sum_{n=0}^{\infty}{(s_n\pi+r_n)x^n}$$ where computing the first few terms suggests that $r_n$, $s_n$ are rational.

Indeed, we have $\frac{\pi/4-\tan^{-1}(\sqrt{1-x})}{\sqrt{1-x}}=\frac{g(x)-f(x)}{2}=\sum{r_nx^n}$, and $\frac{1/4}{\sqrt{1-x}}=\frac{g(x)+f(x)}{2\pi}=\sum{s_nx^n}$.

Using $\frac{d}{dx}[\tan^{-1}(\sqrt{1-x})]=-\frac{1}{2(2-x)\sqrt{1-x}}$, we can use the power series for $\frac{1}{2-x}$ and $\frac{1}{\sqrt{1-x}}$ to calculate the power series for $(\pi/4-\tan^{-1}(\sqrt{1-x}))$, which consists of only rational coefficients. Combining this with the power series for $\frac{1}{\sqrt{1-x}}$ gives: $$r_n=\frac{1}{2}\sum_{m=0}^{n-1}{\sum_{k=0}^{n-m-1}{\frac{{{2k}\choose{k}}{{2m}\choose{m}}}{(n-m)\cdot2^{k+m+n}}}}$$ We easily get $s_n=\frac{1}{2^{2n+2}}{2n\choose{n}}$.

Now, because no branch of $f(z)$ has singularities anywhere (the apparent singularity at $z=1$ is removable), the coefficients of the power series of $f$ must tend to zero.

Hence $\lim_{n\to\infty}{\frac{r_{n+1}}{s_{n+1}}}=\pi$, and the desired limit follows after simplifying.


1) It is easily shown that: $$s_n\pi-r_n=\int_{0}^{\pi/4}{\sin^{2n}{\theta}\,d\theta},\qquad{s}_n\pi+r_n=\int_{\pi/4}^{\pi/2}{\sin^{2n}{\theta}\,d\theta}$$

2) The above proof actually shows: $$\lim_{n\to\infty}{\left(1+\frac{1}{2n+1}\right)\sum_{m=0}^{n}{\sum_{k=0}^{n-m}{\left[\frac{2^{n-m-k}}{n-m+1}\,\frac{{{2k}\choose{k}}{{2m}\choose{m}}}{{{2n}\choose{n}}}\right]}}}=\pi$$ which converges much faster than the given limit.

  • 3
    $\begingroup$ Holy smokes.... $\endgroup$
    – Barron
    Dec 31, 2018 at 19:01

1 Answer 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.


  • 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).


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