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(1/2k)}{2k(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*}