Evaluate $\lim_{ n \to \infty} \sum_{r=2n+1}^{3n} \frac{n}{r^2-n^2}$ 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?
 A: $$\sum_{r=2r+1}^{3n}\dfrac{n^2}{r^2-n^2}$$
$$=\sum_{r=2r+1}^{3n}\dfrac1{(r/n)^2-1}$$
$$=\sum_{r=1}^{3n}\dfrac1{(r/n)^2-1}-\sum_{r=1}^{2n}\dfrac1{(r/n)^2-1}$$
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 $\lim_{n \to \infty }\left(\frac{1}{{n\sqrt{{n^2} + 1}}}+\frac{2}{{n\sqrt{{n^2}+4}}}+\cdots+\frac{n}{{n\sqrt{{n^2}+{n^2}}}}\right)$
A: Good start, but the limit in your final expression should be to the left. You can then write it as a Riemann sum:
$$
S=\lim_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{(2+(k+1)/n)^2-1}=\int_0^1\frac{1}{(2+x)^2-1}\,dx=\cdots=\frac{1}{2}\ln\frac{3}{2}.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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 $\ds{n \to \infty}$.

