I'm trying to find the sum of this series, but I'm not sure how to go about doing that. Based on the Direct Comparison Test, I've found that this function converges, but I don't know how to find it's sum. Any help would greatly be appreciated.

Rusty $$\sum_{n=1}^\infty \dfrac6{n(n+3)}.$$

  • 1
    $\begingroup$ Hint: Partial fractions. $\endgroup$ – Zircht May 17 '14 at 14:43


Use partial fraction decomposition to get $$\frac{6}{n (n+3)}=2(\frac{1}{n} -\frac{1}{n+3})$$ and check if, by chance, they do not telescope.


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