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 $$ \sum_{k=0}^n\frac{\binom{2k}{k}}{2^{2k}}=\frac{2n+1}{2^{2n}}\binom{2n}{n}. $$

Companion to Concrete Mathematics, IIRC) handy, but isn't there a variant of Hadamard convolution that produces the function $f(x) = \sum_n a_nb_nx^n$ from the two functions $g(x)=\sum_n a_nx^n$ and $h(x)=\sum_n b_nx^n$? $\endgroup$ – Steven Stadnicki May 16 '14 at 18:511more comment