Prove that $\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}\sim\frac1{2n}$
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$$n\sum_{i=1}^{n}\frac{1}{(n+i)^2}\to\frac{1}{2}$$