<s>[INST] Combinatorial identity involving the square of $\binom{2n}{n}$ Is there any closed formula for 
$$
\sum_{k=0}^n\frac{\binom{2k}{k}^2}{2^{4k}}
$$
?
This sum of is made out of the square of terms $a_{k}:=\frac{\binom{2k}{k}}{2^{2k}}$ 
I have been trying to verify that $$
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{\left(\sum_{j=0}^k a^2_{j}\right)}{(2k+1)(2k+2)}\right] -\frac{1}{2}{\sum_{k=0}^na^2_{k}}=\frac{1}{2\pi},
$$
which seems to be true numerically using Mathematica.
The question above is equivalent to finding some formula for $$b_{n}:=\frac{1}{2^{2n}}\sum_{j=0}^n\frac{\binom{2n+1}{j}}{2n+1-j}.$$ This is because one can verify that 
$$(2n+1)b_n=2nb_{n-1}+a_n,\qquad a_{n+1}=\frac{2n+1}{2n+2}a_n,$$
and combining these two we get
$$(2n+2)a_{n+1}b_{n}-(2n)a_nb_{n-1}=a_n^2$$
Summing we get
$$\sum_{k=0}^na_k^2=(2n+1)a_nb_n.$$
I also know that 
$$
\frac{\binom{2n}{n}}{2^{2n}}=\binom{-1/2}{n},
$$
so that
$$
\sum_{k=0}^{\infty}\frac{\binom{2k}{k}}{2^{2k}}x^k=(1-x)^{-1/2},\quad |x|<1.
$$
I have also seen the identity
$$
\sum_{k=0}^n\frac{\binom{2k}{k}}{2^{2k}}=\frac{2n+1}{2^{2n}}\binom{2n}{n}.
$$
 [/INST] The limit I want to verify is 
$$
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{\left(\sum_{j=0}^k a^2_{j}\right)}{(2k+1)(2k+2)}\right] -\frac{1}{2}{\sum_{k=0}^na^2_{k}}=\frac{1}{2\pi}
$$
For this it is sufficient to prove that the above expression under the limit is bounded above. I know this is true because of the original problem this limit is coming from, but I do not have a short prove. Then, given that such expression is bounded, we can argue as follows: using summation by parts we see that $$
\sum_{k=0}^{n-1}\left(\frac{1}{2k+1}-\frac{1}{2k+3}\right)\sum_{j=0}^ka^2_{j}=\sum_{k=0}^n\frac{a^2_{k}}{2k+1}-\frac{1}{2n+1}\sum_{k=0}^na^2_{k}
$$ and so we can write the limit as
\begin{align*}
\lim_{n\to\infty} (2n+1)\left[\frac{\pi}{4}-\frac{1}{2}\sum_{k=0}^n\frac{a^2_{k}}{2k+1}-\sum_{k=0}^{n-1}\frac{\sum_{j=0}^ka^2_{j}}{(2k+1)(2k+2)(2k+3)}\right]
\end{align*}
Since the limit of the bracket must be zero in under for the whole expression to remain bounded above, $\pi/4$ must equal the series (sum from $k=0$ up to $\infty$) and since 
\begin{align*}
a_{n}:=\frac{1}{2^{2n}}&\binom{2n}{n}= \frac{\Gamma(1/2)\Gamma(n+1/2)}{\pi\Gamma(n+1)}=  \frac{1+O(1/n)}{\sqrt{\pi n}}
\end{align*} 
the limit becomes 
\begin{align*}
&\lim_{n\to\infty}(2n+1)\left[\frac{1}{2}\sum_{k=n+1}^\infty\frac{a^2_{k}}{2k+1}+\sum_{k=n}^{\infty}\frac{\sum_{j=0}^ka^2_{j}}{(2k+1)(2k+2)(2k+3)}\right]\\
=&\lim_{n\to\infty} \frac{(2n+1)}{2\pi}\sum_{k=n+1}^\infty\frac{1+O(1/k)}{k(2k+1)}+ (2n+1)O\left(\sum_{k=n}^{\infty}\frac{\sum_{j=0}^k\frac{1}{j}}{(2k+1)(2k+2)(2k+3)}\right)\\
=&\frac{1}{2\pi}.
\end{align*}
</s>