Evaluate $\lim_{ n \to \infty} \sum_{r=2n+1}^{3n} \frac{n}{r^2-n^2}$

Question

Evaluate $$S=\lim_{ n \to \infty} \sum_{r=2n+1}^{3n} \frac{n}{r^2-n^2}$$

I have used change of variable $k=r-(2n+1)$

Then

$$S=\frac{1}{n} \times \lim_{ n \to \infty} \sum_{k=0}^{n-1} \frac{n^2}{(2n+1+k)^2-n^2}$$

Then how can we continue?

Answer 1

$$\sum_{r=2r+1}^{3n}\dfrac{n^2}{r^2-n^2}$$

$$=\sum_{r=2r+1}^{3n}\dfrac1{(r/n)^2-1}$$

$$=\sum_{r=1}^{3n}\dfrac1{(r/n)^2-1}-\sum_{r=1}^{2n}\dfrac1{(r/n)^2-1}$$

In the first put $3n=m$ and $2n=t$ in the second and use

The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

OR

Evaluate $\lim_{n \to \infty }\left(\frac{1}{{n\sqrt{{n^2} + 1}}}+\frac{2}{{n\sqrt{{n^2}+4}}}+\cdots+\frac{n}{{n\sqrt{{n^2}+{n^2}}}}\right)$

Answer 2

Good start, but the limit in your final expression should be to the left. You can then write it as a Riemann sum: $$ S=\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{(2+(k+1)/n)^2-1}=\int_0^1\frac{1}{(2+x)^2-1}\,dx=\cdots=\frac{1}{2}\ln\frac{3}{2}. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} S & \equiv \lim_{n \to \infty}\sum_{r = 2n + 1}^{3n}{n \over r^{2} - n^{2}} = {1 \over 2}\lim_{n \to \infty}\pars{\sum_{r = 2n + 1}^{3n}{1 \over r - n} - \sum_{r = 2n + 1}^{3n}{1 \over r + n}} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\pars{\sum_{r = n + 1}^{2n}{1 \over r} - \sum_{r = 3n + 1}^{4n}{1 \over r}} = {1 \over 2}\lim_{n \to \infty}\bracks{\pars{H_{2n} - H_{n}} - \pars{H_{4n} - H_{3n}}} \\[5mm] & = {1 \over 2}\lim_{n \to \infty}\braces{\bracks{\ln\pars{2n} - \ln\pars{n}} - \bracks{\ln\pars{4n} -\ln\pars{3n}}} = {1 \over 2}\bracks{\ln\pars{2} - \ln\pars{4 \over 3}} \\[5mm] & = \bbx{{1 \over 2}\ln\pars{3 \over 2}} \approx 0.2027 \end{align}

Note that $\ds{{\pars{H_{2n} - H_{n}} - \pars{H_{4n} - H_{3n}} \over 2} \sim {1 \over 2}\ln\pars{3 \over 2} + {5 \over 48}\,{1 \over n}}$ as $\ds{n \to \infty}$.