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.
Best Answer
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$$n\sum_{i=1}^{n}\frac{1}{(n+i)^2}\to\frac{1}{2}$$