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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.

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    $\begingroup$ What is wrong with assuming the value of the integral? $\endgroup$ – Math1000 Nov 21 '19 at 19:50
  • $\begingroup$ Within the context of the integral problem, it would be circular - I wanted to see if I could prove the value of the sum without a priori knowing it was a Riemann sum. Also, it seems like a pretty easy asymptote so I'd have thought there would be a nice proof. $\endgroup$ – Jam Nov 21 '19 at 19:53
  • $\begingroup$ I've amended the question to include that I'd like to see if there's another way of showing the relationship without knowing that it's a Riemann sum for the above reason. $\endgroup$ – Jam Nov 21 '19 at 19:55
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    $\begingroup$ Approximate $\frac{1}{k^2}$ with something that telescopes, like $\frac{1}{k^2 - \frac{1}{4}}$ or $\frac{1}{k(k\pm 1)}$ and show the difference doesn't matter. $\endgroup$ – Daniel Fischer Nov 21 '19 at 19:57
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    $\begingroup$ @DanielFischer Thank you Daniel. $\endgroup$ – Jam Nov 21 '19 at 19:59
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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$$n\sum_{i=1}^{n}\frac{1}{(n+i)^2}\to\frac{1}{2}$$

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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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  • $\begingroup$ Perfect. This is what I was looking for. Thank you, user. $\endgroup$ – Jam Nov 21 '19 at 20:01
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    $\begingroup$ The Euler–Maclaurin one? That's kind of weird if you want to avoid integrals. $\endgroup$ – Daniel Fischer Nov 21 '19 at 20:01
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    $\begingroup$ @Jam Nice to may help you! $\endgroup$ – user Nov 21 '19 at 20:03
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    $\begingroup$ @DanielFischer You're right; I hadn't thought of that. $\endgroup$ – Jam Nov 21 '19 at 20:05
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Yet another way: since $\frac{1}{x^2}$ is convex on $\mathbb{R}^+$, the Hermite-Hadamard inequality ensures $$ \frac{1}{2n+1}-\frac{1}{(2n+1)(4n-1)}=\int_{n+1/2}^{2n-1/2}\frac{dx}{x^2}\geq\sum_{k=1}^{n}\frac{1}{(n+k)^2}\geq \int_{n+1}^{2n}\frac{dx}{x^2}=\frac{1}{2n}-\frac{1}{n(n+1)}. $$

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  • $\begingroup$ Sure, but that assumes the integral of $1/x^2$ and, as the question points out, if you know that, you can just use the relation of the sum to a Reimann sum for $\int_1^2dx/x^2$ $\endgroup$ – Thomas Andrews Dec 3 '19 at 8:36
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffc]{\sum_{i = 1}^{n}{1 \over \pars{n + i}^{2}} \sim {1 \over 2n}}:\ {\large ?}}$.


\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, $$ 0 < 2\int_{n}^{2n}{\braces{x} \over x^{3}}\,\dd x < 2\int_{n}^{2n}{\dd x \over x^{3}} = {3 \over 4n^{2}} <<< {1 \over 2n} \quad \mbox{as}\ n \to \infty $$

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