# Ask Uncle Colin: An Infinite Sum

Dear Uncle Colin

I’ve been asked to find $\sum_3^\infty \frac{1}{n^2-4}$. Obviously, I can split that into partial fractions, but then I get two series that diverge! What do I do?

- Which Absolute Losers Like Infinite Series?

Hi, WALLIS, and thanks for your message!

Hey! I’m an absolute loser who likes infinite series, thank you very much!

As you say, you’ve split the sum into partial fractions: $\frac{1}{(n-2)(n+2)} = \frac{1}{4(n-2)} - \frac{1}{4(n+2)}$.

If you write the first few terms out, you get $ \left( \frac{1}{4\times1} - \frac{1}{4\times 5}\right) + \left( \frac{1}{4\times2} - \frac{1}{4\times 6}\right) + \left( \frac{1}{4\times3} - \frac{1}{4\times 7}\right) + \left( \frac{1}{4\times4} - \frac{1}{4\times 8}\right) + \left( \frac{1}{4\times5} - \frac{1}{4\times 9}\right) + …$

The $\frac{1}{4\times 5}$ terms - and, in fact, all of the terms after it - occur as both positive and negative, so they disappear. You’re left with $\frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{16} = \frac{12+6+4+3}{48} = \frac{25}{48}$.

Hope that’s made you like infinite series a little more! I think they’re rather neat - especially this trick of the *telescoping sum*.

- Uncle Colin