Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$

Since: $$\frac{1}{(2k-1)^2+\pi^2}=\int_{0}^{+\infty}\frac{\sin(\pi x)}{\pi}e^{-(2k-1)x}\,dx $$ and: $$\sum_{k=1}^{n}\binom{2n-1}{n-k}e^{-(2k-1)x}=\binom{2n-1}{n-1}\cdot\sum_{h=0}^{+\infty}\left(e^{-(2h+1)x}\prod_{k=0}^{h}\frac{n-k}{n+k}\right),$$ $$\lim_{n\to +\infty}\frac{\sqrt{n}}{4^n}\binom{2n-1}{n-1}=\frac{1}{2\sqrt{\pi}},$$ the value of the limit is given by: $$L=\frac{1}{2\pi^{3/2}}\int_{0}^{+\infty}\sum_{h=0}^{+\infty}\left(e^{-(2h+1)x}\sin(\pi x)\prod_{k=0}^{h}\frac{n-k}{n+k}\right)dx.$$ As long as $n\to +\infty$, the products appearing in the previous line get closer and closer to $1$.

By the dominated convergence theorem,

$$ L = \frac{1}{2\sqrt{\pi}}\sum_{h=0}^{+\infty}\frac{1}{(2h+1)^2+\pi^2}=\frac{1}{8\sqrt{\pi}}\tanh\frac{\pi^2}{2},$$

where the last identity follows by considering the logarithmic derivatives of the Weierstrass product for the $\cosh$ function.

This answer matches up with Jack D'Aurizio's at $(3)$, but I thought it was distinct enough not to delete it.

---

The second derivative of $\log\binom{2n-1}{n-k}$ between $k=0$ and $k=1$ is $-\frac2n$. This means that a good approximation for $\binom{2n-1}{n-k}$ is $\frac{2^{2n-1}}{\sqrt{\pi n}}e^{-(k-\frac12)^2/n}$. For example at $n=100$,

$\hspace{3.4cm}$

Since equation $(9)$ in this answer is valid for all $z\in\mathbb{C}$, we have $$ \sum_{k=1}^\infty\frac{1}{k^2+z^2}=\frac1{2z}\left[\pi\coth(\pi z)-\frac1z\right]\tag{1} $$ Thus, we can compute $$ \begin{align} &\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}\\ &=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\frac{2^{2n-1}}{\sqrt{\pi n}}\sum_{k=1}^n\frac{e^{-(k-\frac12)^2/n}}{(2k-1)^2+\pi^2}\tag{2}\\ &=\frac1{2\sqrt\pi}\sum_{k=1}^\infty\frac1{(2k-1)^2+\pi^2}\tag{3}\\ &=\frac1{2\sqrt\pi}\left(\sum_{k=1}^\infty\frac1{k^2+\pi^2}-\sum_{k=1}^\infty\frac1{4k^2+\pi^2}\right)\tag{4}\\ &=\frac1{2\sqrt\pi}\left(\color{#C00000}{\sum_{k=1}^\infty\frac1{k^2+\pi^2}}\color{#00A000}{-\frac14\sum_{k=1}^\infty\frac1{k^2+\pi^2/4}}\right)\tag{5}\\ &=\frac1{2\sqrt\pi}\left(\color{#C00000}{\frac12\coth(\pi^2)-\frac1{2\pi^2}}\color{#00A000}{-\frac14\coth(\pi^2/2)+\frac1{2\pi^2}}\right)\tag{6}\\ &=\frac1{8\sqrt\pi}\tanh(\pi^2/2)\tag{7} \end{align} $$ Explanation:
$(2)$: approximate the binomial by the appropriate normal distribution
$(3)$: cancel terms and apply Dominated Convergence (or Monotone Convergence, either works)
$(4)$: the sum over the odds is the full sum minus the evens
$(5)$: prepare to apply $(1)$
$(6)$: apply $(1)$
$(7)$: cancel terms and use $2\coth(2x)-\coth(x)=\tanh(x)$

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 $$ \lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}=\frac1{8\sqrt\pi}\tanh(\pi^2/2)\tag{11} $$