Consider the integral $$I=\int_0^1\int_0^1\frac{zdzdt}{(1-zt)(1-z(1-t))}=\sum_{k,j\geqslant 0} \int_0^1\int_0^1 z^{k+j+1}t^k(1-t)^jdzdt=\\
\sum_{k,j\geqslant 0} \frac1{k+j+2}\cdot \frac{k!j!}{(k+j+1)!},$$
that is your sum (denote $n=k+j$). For evaluating the integral, we first integrate by $t$, get $-2\log(1-z)/(2-z)$. Now denote $1-z=x$ and integrate $2\log x/(1+x)=2\log x(1-x+x^2-x^3+\dots)$ using $\int_0^1 -\log x\cdot x^m dx=\frac1{(m+1)^2}$. We get $2(1-1/4+1/9-1/16+\dots)=\pi^2/6$.

As for your second sum, it has indeed the same value which may be obtained similarly. We consider the integral
$$
\int_0^1\int_0^1 \frac1x\left(\frac1{1-y\frac{1+x}2}-\frac1{1-y\frac{1-x}2}\right)dxdy=\\ \sum_n \int_0^1y^n dy\cdot 2^{1-n}\int_0^1\frac{(1+x)^n-(1-x)^n}{2x}dx,
$$
that is your sum (expand $(1+x)^n-(1-x)^n=2\sum \binom{n}{2k-1}x^{2k-1}$ and integrate).

If we integrate first in $x$, we get $-2\log(1-y)/(2-y)$, the same thing as in the first sum.

"By the comparison test, the series converges."$\endgroup$ – Alex M. Jun 3 '18 at 17:24