Prove that $\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}\sim\frac1{2n}$ I would like a proof of the asymptotic relationship $$\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}\sim\frac1{2n}$$
without assuming that the sum is a Riemann sum.
This problem arose from Question 1909556, which asks about the Riemann sum of $\int_1^2\frac1{x^2}\ \mathrm{d}x=\lim_{n\to\infty}\frac1n\sum_{i=0}^n\frac{n^2}{(n+i)^2}=\frac12$. It features a nebulous clue that $\lim_{n\to\infty}\frac{\sum_{i=1}^{2n}\frac1{i^2}-\sum_{i=1}^{n}\frac1{i^2}}{1/n}=\frac12$. I can't figure out how this clue works but one way of showing it could be with the asymptotic relationship is true, and from calculations, it seems to work. But I can't find any feasible way of proving it without assuming the value of the integral.
I would like to avoid assuming that the sum is simply the integral so that I can prove the integral from the sum. It also seems like a fairly simple relationship, so I would imagine there could be a nice proof.

We have that
$$\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}=\sum_{i=n+1}^{2n}\frac{1}{i^{2}}=\sum_{i=1}^{2n}\frac{1}{i^{2}}-\sum_{i=1}^{n}\frac{1}{i^{2}}$$
then we can use the result indicated here


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*What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?
