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
$$
\sum_{k=0}^n\frac{\binom{2k}{k}}{2^{2k}}=\frac{2n+1}{2^{2n}}\binom{2n}{n}.
$$
 A: Mathematica says:
$$\sum_{k=0}^n \binom{2k}{k}^2 x^k = 
\frac{2 K(16 x)}{\pi }-\binom{2 (n+1)}{n+1}^2 x^{n+1} \,
   _3F_2\left(1,n+\frac{3}{2},n+\frac{3}{2};n+2,n+2;16 x\right).$$
(the previous answer had "pilot error", as Christian noticed), but in my defence, the OP does not square the binomial in most of his question.
Further 
$$2^{2 n} b_n = \frac{n \, _2F_1(-2 n-1,-2 n-1;-2 n;-1)-(2 n+1) \binom{2 n+1}{n+1} \,
   _3F_2(1,-n,-n;1-n,n+2;-1)}{n (2 n+1)}.$$
A: 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*}
A: (Not yet an answer, but too long for a comment).
The expression $\frac{\binom{2n}{n}^2}{16^n}$ occur in random walks on
lattice grids in the plane, (going back to Polya).
See e.g. http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter12.pdf
and Doyle and Snell:
http://arxiv.org/pdf/math/0001057
section 2.3.3.
The fact that the infinite sum diverges is Polya's result
that the random walk will almost surely return.
Mathematica  Sum[Binomial[2 k, k]^2/16^k, {k, 0, n}]
gives an expression which is hard to read/interpret,
but links to some difference equation,
Maybe this helps...
DifferenceRoot[
  Function[{\[FormalY], \[FormalN]}, {(1 +
         2 \[FormalN])^2 \[FormalY][\[FormalN]] + (-5 -
         12 \[FormalN] - 8 \[FormalN]^2) \[FormalY][1 + \[FormalN]] +
      4 (1 + \[FormalN])^2 \[FormalY][2 + \[FormalN]] ==
     0, \[FormalY][0] == 0, \[FormalY][1] == 1}]][1 + n]

Also of interest is maybe:
Sum[Binomial[2 k, k]^2 x^k, {k, 0, n}]

(2 EllipticK[16 x])/\[Pi] - 
 x^(1 + n) Binomial[2 (1 + n), 
   1 + n]^2 HypergeometricPFQ[{1, 3/2 + n, 3/2 + n}, {2 + n, 2 + n}, 
   16 x]

(if anybody can display/interpret this in a better form, please do)
A: The generating function for $b_n$ can be expressed as
\begin{split}
\sum_{n\geq 0} b_n z^{2n} &= 
\frac{2}{z(z+2)}\cdot\log\frac{1+z-\sqrt{1-z^2}}z \\
&+ \frac{2}{z(z-2)}\cdot\log\frac{\sqrt{1-z^2}-1+z}z\\
&- \frac{4}{(z-2)(z+2)}\cdot\log\frac{1+\sqrt{1-z^2}}2.
\end{split}
Just in case, here is numerical verification in Maple:
> seq( add( binomial(2*n+1,j) / (2*n+1-j), j=0..n ) / 2^(2*n), n=0..6 );
                 11  287  5989   114859  4215377   188220881
              1, --, ---, -----, ------, --------, ----------
                 24  960  26880  645120  28385280  1476034560

> series( 2/(z+2)/z*ln( (1+z-sqrt(1-z^2))/z ) +  2/(z-2)/z*ln( (sqrt(1-z^2)-1+z)/z ) - 4/(z-2)/(z+2)*ln( (1+sqrt(1-z^2))/2 ), z, 14);
    11  2   287  4   5989   6   114859  8   4215377   10   188220881   12      14
1 + -- z  + --- z  + ----- z  + ------ z  + -------- z   + ---------- z   + O(z  )
    24      960      26880      645120      28385280       1476034560

