A more elementary approach.
$$\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{(n+i)(n+i-1)}$$ because the sum telescopes to $\frac{1}{n}-\frac{1}{2n}.$
So:
$$\begin{align}\frac{1}{2n}-\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}&=\sum_{i=1}^{n}\left(\frac{1}{(n+i)(n+i-1)}-\frac{1}{(n+i)^2}\right)\\
&=\sum_{i=1}^{n}\frac{1}{(n+i)^2(n+i-1)}\tag{1}\\&<\sum_{i=1}^{n}\frac{1}{n^3}\\&=\frac{1}{n^2}
\end{align}$$
Also, the value at (1) is positive. So we have:
$$0<\frac{1}{2}-n\sum_{i=1}^{n}\frac{1}{(n+i)^2}<\frac{1}{n}$$ and hence$$n\sum_{i=1}^{n}\frac{1}{(n+i)^2}\to\frac{1}{2}$$
I should consider
$$f(x)=\sum_{n=0}^\infty\frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}x^n=\sum_{n=0}^\infty a_n\,x^n$$ Expand and separate odd and even terms, we have
$$S_1(x)=\sum_{n=0}^\infty a_{2n}\,x^{2 n}=\frac{2}{\sqrt{\pi }}+\frac{4 x^2 }{9 \sqrt{\pi }}\,
_3F_2\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x^2
\right)$$
$$S_2(x)=\sum_{n=0}^\infty a_{2n+1}\,x^{2 n+1}=\frac{\sqrt{\pi }}{x}\left(\log (2)-\log \left(1+\sqrt{1-x^2}\right)\right)$$
Making $x=\pm 1$ to cover the two cases
$$S_1(\pm 1)=\frac{4 C}{\sqrt{\pi }}\qquad \text{and} \qquad
S_2(\pm 1)=\pm \sqrt{\pi } \log(2)$$ and hence the results.
Edit (to clarify)
Write
$$a_{2n}=\frac{2}{\sqrt{\pi }}\frac {b_n}{c_n}$$ The $b_n$
$$\{1,2,8,16,128,256,1024,2048,32768,65536\} $$
are the denominators of
$$\frac 1 {4^n} \binom{2 n}{n}$$ and the $c_n$
$$\{1,9,75,245,2835,7623,39039,96525,1859715,4387955\}$$ are the denominators of the series expansion of
$$\,
_3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};x
\right)$$
Write
$$a_{2n+1}=\frac{\sqrt{\pi }}4\frac {d_n}{e_n}$$
The $d_n$
$$\{1,3,5,35,63,77,429,6435,12155,46189\}$$ are the numerators of the series expansion of
$$\cosh ^{-1}(x)-\log (2 x)$$ and the $e_n$
$$\{1,8,24,256,640,1024,7168,131072,294912,1310720\}$$ had to be worked !
Best Answer
If $x=0$ the result is false since the sum is equal to $1$ (the term with $n=0$ remains). So assume that $x>0$. We have, by the Newton binomial theorem, that$$\frac{1}{x!}\sum_{n\geq0}\dbinom{x}{n}\left(-1\right)^{n}=\frac{\left(1-1\right)^{x}}{x!}=0$$ and so the thesis.