# Thread: Stuck on evaluating a series

1. ## Stuck on evaluating a series

How would evaluate the following series?

$\sum _{i=3}^{\infty }2\,{\frac {1}{ \left( i-2 \right) \left( i+2
\right) }}$

I can get my calculator to evaluate this, but I am stuck when trying to calculate it manually. Is this a telescoping series where I would need to use partial fractions? I can't get it to work out that way.

2. Originally Posted by RU2010
How would evaluate the following series?

$\sum _{i=3}^{\infty }2\,{\frac {1}{ \left( i-2 \right) \left( i+2
\right) }}$

I can get my calculator to evaluate this, but I am stuck when trying to calculate it manually. Is this a telescoping series where I would need to use partial fractions? I can't get it to work out that way.
It is not quite a telescoping series, but do use partial fractions and write out the first few terms and you will see what actually cancels out.

CB

3. These are nearly always able to be evaluated as a telescopic series after a partial fraction decomposition.

$\frac{A}{i - 2} + \frac{B}{i + 2} = \frac{1}{(i - 2)(i + 2)}$

$\frac{A(i + 2) + B(i - 2)}{(i - 2)(i + 2)} = \frac{1}{(i - 2)(i + 2)}$

$A(i + 2) + B(i - 2) = 1$

$Ai + 2A + Bi - 2B = 1$

$(A + B)i + 2A - 2B = 0i + 1$.

$A+B=0$ and $2A - 2B = 1$.

Substituting $B = -A$ into the second equation gives

$2A + 2A = 1$

$4A = 1$

$A = \frac{1}{4}$.

Therefore $B= -\frac{1}{4}$.

So finally we have

$\frac{1}{(i - 2)(i + 2)} = \frac{1}{4(i - 2)} - \frac{1}{4(i + 2)}$.

Therefore our series is

$2\sum_{i = 3}^{\infty}\frac{1}{(i - 2)(i + 2)} = 2\lim_{n \to \infty}\sum_{i = 3}^{n}\left[\frac{1}{4(i - 2)} - \frac{1}{4(i + 2)}\right]$

$= \frac{1}{2}\lim_{n \to \infty}\sum_{i = 3}^{n}\left(\frac{1}{i - 2} - \frac{1}{i + 2}\right)$

$= \frac{1}{2}\lim_{n \to \infty}\left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{2} - \frac{1}{6}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \left(\frac{1}{5} - \frac{1}{9}\right) + \left(\frac{1}{6} - \frac{1}{10}\right) + \left(\frac{1}{7} - \frac{1}{11}\right) + \dots + \left(\frac{1}{n - 2} - \frac{1}{n + 2}\right)$

$= \frac{1}{2}\lim_{n \to \infty}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{n - 1} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}\right)$ since everything cancels except the first and last four terms

$= \frac{1}{2}\lim_{n \to \infty}\left(\frac{25}{12} - \frac{1}{n - 1} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}\right)$

$= \frac{1}{2}\left(\frac{25}{12}\right)$

$= \frac{25}{24}$.

4. Got it! Thanks!