<s>[INST] Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$ What tools would you recommend me for computing the limit below?
$$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$
As soon as any useful idea comes to mind I'll make the proper update with the new findings.
 [/INST] After I posted my earlier answer, I realized that this can be handled in a much simpler way.
Pick an $\epsilon\gt0$.

Most of the series is contained in a finite sum
This identity is proven in my earlier answer:
$$
\sum_{k=1}^\infty\frac1{(2k-1)^2+\pi^2}=\frac14\tanh(\pi^2/2)\tag{1}
$$
Since the series in $(1)$ converges, we know that there is a $K$ so that
$$
0\le\sum_{k=K+1}^\infty\frac1{(2k-1)^2+\pi^2}=\frac14\tanh(\pi^2/2)-\sum_{k=1}^K\frac1{(2k-1)^2+\pi^2}\le\epsilon\tag{2}
$$

Where most of the series is contained, the coefficients are almost equal
Now, using $(9)$ from this answer, we have
$$
\frac{\sqrt{n}}{4^n}\binom{2n-1}{n}=\frac{\sqrt{n}}{4^n}\frac12\binom{2n}{n}\sim\frac{\sqrt{n}}{4^n}\frac12\frac{4^n}{\sqrt{\pi n}}=\frac1{2\sqrt{\pi}}\tag{3}
$$
Therefore, there is an $M$ so that for $n\ge M$,
$$
\frac1{2\sqrt{\pi}}(1-\epsilon)\le\frac{\sqrt{n}}{4^n}\binom{2n-1}{n}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{4}
$$
Consider that
$$
\binom{2n-1}{n-k}=\binom{2n-1}{n}\frac{n-1}{n+1}\frac{n-2}{n+2}\cdots\frac{n-k+1}{n+k-1}\tag{5}
$$
Since we can choose an $N$ so that for all $n\ge N$ and $k\le K$,
$$
1-\epsilon\le\frac{n-1}{n+1}\frac{n-2}{n+2}\cdots\frac{n-k+1}{n+k-1}\le1\tag{6}
$$
we can combine $(4)$, $(5)$, and $(6)$ to get that for all $n\ge\max(M,N)$ and $k\le K$,
$$
\frac1{2\sqrt{\pi}}(1-\epsilon)^2\le\frac{\sqrt{n}}{4^n}\binom{2n-1}{n-k}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{7}
$$

The coefficients are bounded everywhere
Notice that $(4)$ and $(5)$ also tell us that for all $n\ge M$,
$$
\frac{\sqrt{n}}{4^n}\binom{2n-1}{n-k}\le\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{8}
$$

Put it all together
For $n\ge M$, we have from $(2)$ and $(8)$,
$$
\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}
\le\frac14\tanh(\pi^2/2)\frac1{2\sqrt{\pi}}(1+\epsilon)\tag{9}
$$
For $n\ge\max(M,N)$, we have from $(2)$ and $(7)$,
$$
\frac{\sqrt{n}}{4^n}\sum_{k=1}^K\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}
\ge\left(\frac14\tanh(\pi^2/2)-\epsilon\right)\frac1{2\sqrt{\pi}}(1-\epsilon)^2\tag{10}
$$
Since $\epsilon\gt0$ was arbitrary, we have
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}=\frac1{8\sqrt\pi}\tanh(\pi^2/2)\tag{11}
$$
</s>