I'm working on determining whether the following series converges or diverges:

$$\sum_{n=1}^\infty \frac{9}{n(n+3)}$$

I used partial fraction decomposition to turn it into this:

$$\sum_{n=1}^\infty (\frac{3}{n}-\frac{3}{n+3})$$

And I know that if I use my summation properties, I can make these two separate sums and then find whether they converge or diverge individually. However, once I saw the partial fractions, I realized one of them was a harmonic series, so if one part of the sum diverges, the rest will too.

Then, I also realized that it looked like a telescoping series, and when I wrote out the terms, I found that the series summed to $\frac{11}{2}$. My answer key also had that, although, now I'm confused as to why the logic of the harmonic series results in a different answer. Am I missing something here? Any help is appreciated!

  • 2
    $\begingroup$ What you noticed is exhibited by s simpler sum, $\sum_{n=1}^\infty (n-n).$ $\endgroup$
    – coffeemath
    Commented Mar 16 at 4:53

1 Answer 1


When you split it as a partial fraction, a term is getting subtracted from the harmonic series. That is what makes it converge.

Alternatively, use the Comparision Test: $$\frac{1}{n(n+3)}<\frac{1}{n^2}$$ The second series is a $p$ series with $p=2$ which means it converges, and hence the given series converges.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .