Xrfgtjtk

Consider the following limit:
$$lim_{ntoinfty}{sum_{m=0}^{n}{sum_{k=0}^{n-m}{left[frac{2^{n-m-k}}{n-m+1},frac{{{2k}choose{k}}{{2m}choose{m}}}{{{2n}choose{n}}}right]}}}=pi$$
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:
$$f(x)=sum_{n=0}^{infty}{(s_npi-r_n)x^n},qquad{g}(x)=sum_{n=0}^{infty}{(s_npi+r_n)x^n}$$
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 $frac{1/4}{sqrt{1-x}}=frac{g(x)+f(x)}{2pi}=sum{s_nx^n}$.

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 $s_n=frac{1}{2^{2n+2}}{2nchoose{n}}$.

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

Notes:

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

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

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}^nsum_{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_nsim pibinom{2n}{n}simsqrt{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}^inftyleft( sum_{k=0}^na_kright)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^ntag{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}^1sum_{m=0}^nsum_{k=0}^{n-m}binom{2k}{k}binom{2m}{m}2^{n-m-k}z^{n-m},dztag{4}\
&=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}sum_{k=0}^{n-m}binom{2k}{k}2^{n-m-k},dztag{5}\
&=int_{0}^1sum_{m=0}^nbinom{2m}{m}z^{n-m}[t^{n-m}]frac{1}{(1-2t)sqrt{1-4t}},dztag{6}\
&=int_{0}^1sum_{m=0}^inftybinom{2n-2m}{n-m}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{7}\
&=int_{0}^inftysum_{m=0}^infty[u^{n-m}]frac{1}{sqrt{1-4u}}z^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{8}\
&=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1sum_{m=0}^infty(zu)^m[t^m]frac{1}{(1-2t)sqrt{1-4t}},dztag{9}\
&=[u^n]frac{1}{sqrt{1-4u}}int_{0}^1frac{1}{(1-2zu)sqrt{1-4zu}},dztag{10}\
&=[u^n]frac{1}{sqrt{1-4u}}left.frac{arctanleft(sqrt{1-4zu}right)}{u}right|_{z=0}^{z=1}tag{11}\
&,,color{blue}{=[u^{n+1}]frac{1}{sqrt{1-4u}}left(-arctanleft(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 $pne -1$.




• 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 $mto n-m$ 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).