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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

O.M
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3 Answers3

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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 $$f(3)=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)$$

user712576
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  • You have to have a certain reputation on the site before you can comment. I don't remember what the minimum is, but I've upvoted your post, so you're on your way. Welcome to MSE. – saulspatz Oct 10 '19 at 18:04
  • Hey thanks, kind of weird they let me answer but not comment. – user712576 Oct 10 '19 at 18:12
  • It's just a matter of making sure you're acculturated to the site. You'll find there are other things you can't do until you get more rep. If you edit a post, your edits will have to be approved before they're posted. You also won't be able to upload images directly. (Include a link to the image, and someone will post it for you.) As you use the site, I think it will become clear why these rules are in place. – saulspatz Oct 10 '19 at 18:18
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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 $$\frac11+\frac12+\frac13.$$

Angina Seng
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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.

Z Ahmed
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