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


*

*What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?
A: Yet another way: since $\frac{1}{x^2}$ is convex on $\mathbb{R}^+$, the Hermite-Hadamard inequality ensures
$$ \frac{1}{2n+1}-\frac{1}{(2n+1)(4n-1)}=\int_{n+1/2}^{2n-1/2}\frac{dx}{x^2}\geq\sum_{k=1}^{n}\frac{1}{(n+k)^2}\geq \int_{n+1}^{2n}\frac{dx}{x^2}=\frac{1}{2n}-\frac{1}{n(n+1)}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[15px,#ffc]{\sum_{i = 1}^{n}{1 \over \pars{n + i}^{2}} \sim
{1 \over 2n}}:\ {\large ?}}$.

\begin{align}
&\bbox[15px,#ffc]{\sum_{i = 1}^{n}{1 \over \pars{n + i}^{2}}} =
\sum_{i = n + 1}^{2n}{1 \over i^{2}} =
\sum_{i = 1}^{2n}{1 \over i^{2}} - \sum_{i = 1}^{n}{1 \over i^{2}}
\\[5mm] & =
\bracks{\zeta\pars{2} - {1 \over 2n} +
2\int_{2n}^{\infty}{\braces{x} \over x^{3}}\,\dd x} -
\bracks{\zeta\pars{2} - {1 \over n} +
2\int_{n}^{\infty}{\braces{x} \over x^{3}}\,\dd x}
\end{align}
In the last line I used a
Zeta Function Identity.
Then,
$$
\bbox[15px,#ffc]{\sum_{i = 1}^{n}{1 \over \pars{n + i}^{2}}} =
{1 \over 2n} - 2\int_{n}^{2n}{\braces{x} \over x^{3}}\,\dd x
$$
However,
$$
0 < 2\int_{n}^{2n}{\braces{x} \over x^{3}}\,\dd x <
2\int_{n}^{2n}{\dd x \over x^{3}} = {3 \over 4n^{2}} <<<
{1 \over 2n} \quad \mbox{as}\ n \to \infty
$$
