Looking for: $\lim_{n \to \infty}\frac{n!}{B_n}\sum_{i=0}^{n}\frac{B_{n-i}B_{i}}{(n-i)!i!}x^{1-i}=F(x)$ Let $n=2k,k=1,2,3,,$
Proposed:
$$\lim_{n \to \infty}\frac{n!}{B_n}\sum_{i=0}^{n}\frac{B_{n-i}B_{i}}{(n-i)!i!}x^{1-i}=F(x)\tag1$$
Where $B_n$ is Bernoulli numbers
$\phi=\frac{1+\sqrt{5}}{2}$
I am looking for $F(x)$, I was able to evaluate for some values of $F(x)$, such as
$$F(3)=\frac{\pi}{\sqrt{3}}$$
$$F(4)=\pi$$
$$F(5)=\frac{\pi\phi^2}{\sqrt{\phi^2+1}}$$
$$F(6)=\pi\sqrt{3}$$
$$F(8)=\pi(1+\sqrt{2})$$
$$F(10)=\pi\sqrt{\phi^3\sqrt{5}}$$
$$F(12)=\pi(2+\sqrt{3})$$
 A: The answer is $\boxed{F(x)=\pi\cot(\pi/x)}$ for $|x|>1$. Let's fix such an $x$ and consider $$f_n=\sum_{k=0}^n\frac{B_k}{k!}\frac{B_{n-k}}{(n-k)!}x^{1-k},\qquad f(t)=\sum_{n=0}^\infty f_n t^n$$ for $|t|$ sufficiently small (in fact, for $|t|<2\pi$); then we're looking for $$F(x)=\lim_{n\to\infty}\frac{(2n)!}{B_{2n}}f_{2n}=-\frac12\lim_{n\to\infty}(-4\pi^2)^n f_{2n}.\tag{*}\label{stophere}$$
Using $\sum_{n=0}^\infty\sum_{k=0}^n=\sum_{k=0}^\infty\sum_{n=k}^\infty$ and $\sum_{n=0}^\infty(B_n/n!)z^n=z/(e^z-1)$, we get $$f(t)=\sum_{k=0}^\infty\frac{B_k}{k!}t^k x^{1-k}\sum_{n=k}^\infty\frac{B_{n-k}}{(n-k)!}t^{n-k}=\frac{t^2}{(e^t-1)(e^{t/x}-1)}.$$
Now $f(2\pi it)$ has simple poles at $t=\pm 1$ with residues $2\pi i/(e^{\pm 2\pi i/x}-1)$. Thus, if $$g(t)=\frac{2\pi i}{(e^{2\pi i/x}-1)(t-1)}+\frac{2\pi i}{(e^{-2\pi i/x}-1)(t+1)},$$ then $f(2\pi it)-g(t)$ is analytic in a neighborhood of $|t|\leqslant 1$. We find $$g(t)=\sum_{n=0}^\infty g_n t^n,\qquad g_n=\begin{cases}\hfill 2\pi i,\hfill & n\text{ is odd}\\-2\pi\cot(\pi/x), & n\text{ is even}\end{cases}$$ so that the analyticity of $f(2\pi it)-g(t)$ (as mentioned above) gives $$\lim_{n\to\infty}\big((2\pi i)^n f_n-g_n\big)=0.$$ It remains to recall \eqref{stophere} and use the value of $g_n$ for even $n$.
