Find the sum of the following series $\sum_{n=1}^\infty \frac{1}{n(n+m)}$

Question

I am suppose to find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+m)}$

But after using partial fractions and getting $ \frac 1m \sum_{n=1}^\infty \frac{1}{n} -\frac{1}{n+m}$

then i couldnt figure out what to do

Answer 1

I'll do $m=3$. Then $m=3$ times the sum is then $$\left(\frac11-\frac14\right) +\left(\frac12-\frac15\right) +\left(\frac13-\frac16\right) +\left(\frac14-\frac17\right)+\cdots. $$ The brackets three apart telescope, leaving just $$\frac11+\frac12+\frac13.$$

Answer 2

For some reason, the site wont allow me to comment. But anyways, a supposed hint would be, if; $$f(m)= \frac 1m \sum_{n=1}^\infty \frac{1}{n} -\frac{1}{n+m}$$ This would just be; (As this is a telescoping sum) $$f(m)=\frac{1}{m}\sum_{n=1}^m \frac{1}{n}$$

Plugging in the value of $m=3$, you get $$f(3)=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)$$

Answer 3

Use the telescopic summation: $$H_x=\lim_{n\rightarrow \infty} \sum_{k=1}^{n} \left(\frac{1}{n}-\frac{1}{n+x} \right)= x \sum_{k=1}^{\infty} \frac{1}{n(n+x)}$$ See

https://en.wikipedia.org/wiki/Harmonic_number

So the required sum is nothing but $\frac{H_m}{m}.$ where $H_m$ are Harmonic numbers.