Hiya guys!

Could someone correct this for me please?Not sure if I went about it right at all!

Determine the sum of the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+2)}$

What I did so far was:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac{1}{1\cdo t{3}}+\frac{1}{2\cdot{4}}+\frac{1}{3\cdot{5}}+...$

The partial sum:

$\displaystyle S_{n}=\frac{1}{1\cdot{3}}+\frac{1}{2\cdot{4}}+\fra c{1}{3\cdot{5}}+...+\frac{1}{(n-2)n}+\frac{1}{n(n+2)}$

Rewritten as:

$\displaystyle (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-{1}{(n+2)})$

(I don't really know why I did this to be honest, I was mostly copying a similar sum from the book, so more than likely gone wrong with that line)

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

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

$\displaystyle S_{n}=\frac{3n^{2}-2n-4}{2n^{2}-4n}$

So...did I go completely wrong??Or where do I go from here to prove it's convergent and what $\displaystyle S_{n}\rightarrow{??}$ is??

Thanks for your help guys!!

Jo