# Evaluate Infinite Series Summation of 1/(k(k+c)) [duplicate]

I am trying to evaluate the infinite series:

$$\sum_{k=1}^{\infty} \frac{1}{k(k+c)}$$

where c is a constant, positive integer.

Is my approach correct by using partial fraction decomposition? Then I break down the series into the form $$\frac{1/c}{k} - \frac{1/c}{k+c}$$ and I get a series of the form (after writing out few terms and noticing some cancellations)

$$\frac{1/c}{1} + \frac{1/c}{2} + \frac{1/c}{3} + ...$$

Is there a way for me to formalize the solution better (if correct), maybe into some kind of compact formula? Otherwise is there a correct/better approach to the problem?

I also used an online infinite series calculator which told me the sum of the series for different values of c. For c = 1,2,3,4 the sum of the infinite series evaluates to $$1, \frac{3}{4}, \frac{11}{18}, \frac{25}{48}$$ respectively. However, I don't really see the pattern here.

Any helps/hints would be much appreciated! Thank you!

## marked as duplicate by Sil, Xander Henderson, stressed out, user3658307, YiFanFeb 24 at 1:57

• Sure, basically you can use approach0.xyz/search , and enter any latex like expression, it will then show you results (it is smarter than full text search, it can find expressions which differ by substitutions etc...). In this case I just searched \sum_{k=1}^{\infty} \frac{1}{k(k+c)} – Sil Feb 23 at 19:31
$$\sum_{n\geq 1}\frac{1}{n(n+m)}=\frac{1}{m}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+m}\right)=\frac{1}{m}\sum_{n=1}^{m}\frac{1}{n}=\frac{H_{m}}{m}$$ where $$H_s$$ is the $$s$$-th harmonic number.