# 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?

## 3 Answers

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

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

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

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

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


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 $\ds{n \to \infty}$.