# Thread: [SOLVED] Finding the Sum of this Series

1. ## [SOLVED] Finding the Sum of this Series

I've attached the problem I'm working on. I know how to do geometric series problems, but don't know how to attack this type of infinite series, and can't really make sense out of what my book tells me to do. I know it involves partial fractions somehow I just don't know what to do exactly.

Could someone help me out or at least clue me in on the procedure?

2. You are right

$\frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2}$

When you sum, all the terms vanish except the two first ones

3. You can find something close to your problem here
http://www.mathhelpforum.com/math-he...ce-series.html

4. You want to first do partial fractions:

$\frac{2}{n(n+2)} = \frac{A}{n}+\frac{B}{n+2}$

Multiplying through by $n(n+2)$ gives you $2 = A(n+2)+Bn = (A+B)n+2A$

Since this is an identity, solve for $A$ and $B$:
$
\left[
\begin{array}{lr}
A+B=0 \\
2A=2
\end{array}
\right]
$

Hence $A = 1$ and $B=-1$.

Now you know that $\frac{2}{n(n+2)} = \frac{1}{n}-\frac{1}{n+2}$

Now we have the sum $\sum_{n=2}^{\infty}\left(\frac{2}{n(n+2)} = \frac{1}{n}-\frac{1}{n+2}\right)$

Let's write out the first several terms: $\frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+...$

The rest of the solution is below, if you need it.
Spoiler:

You see that most of these terms will cancel out. (This is called a telescoping series.) In fact the only two that won't cancel out are $\frac{1}{2}$ and $\frac{1}{3}$.

Thus, $\sum_{n=2}^{\infty}\frac{2}{n(n+2)} = \frac{1}{2}+\frac{1}{3} = \frac{5}{6}$