# Thread: Series correction

1. ## Series correction

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 $\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$

What I did so far was:
$\sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac{1}{1\cdo t{3}}+\frac{1}{2\cdot{4}}+\frac{1}{3\cdot{5}}+...$

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

Rewritten as:

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

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

$S_{n}=\frac{(2)(n-2)(n)+(n-2)(n)+(2)(n)+(2)(n-2)}{(2)(n-2)(n)}$
$S_{n}=\frac{3n^{2}-2n-4}{2n^{2}-4n}$

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

Thanks for your help guys!!
Jo

2. Originally Posted by simplysparklers
Rewritten as:

$(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-{1}{(n+2)})$
(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)
You knew it all along

Always try to find the partial fraction decomposition in these cases:

Here $\frac1{n(n+2)} = \frac12 \left(\frac1{n} - \frac1{n+2}\right)$. So your sum should have read:

$\frac12\left((1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{(n+2)})\right)$

3. Thanks Isomorphism!
Just wondering though, did I still pick the right terms left over from the series?Does it come out then as:
$S_{n}=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{(n-2)}+\frac{1}{n})$?

Which goes to:
$\frac{3n^{2}-2n-4}{4n^{2}-8n}$?

Thanks,
Jo

4. Originally Posted by simplysparklers
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 $\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$

Most likely it is wrong since you did not consider the n+1 term...

But anyway I know a way to do it...

$\sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+2} = \frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+1} - \frac{1}{n+2}$

$\frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} +\frac12 \sum_{n=1}^{\infty} \frac{1}{n+1} - \frac{1}{n+2} = \frac12 \lim_{n \to \infty} 1 - \frac1{n+1} + \frac12 - \frac1{n+2} = \frac34$

EDIT: Your answer is mostly correct since you get the same limit.

5. Hugs for Isomorphism!!! Thank you!!

6. From MacLaurin series:

$\sum\limits_{n\,=\,1}^{\infty }{\frac{1}{n(n+2)}}=\int_{0}^{1}{\left\{ x\sum\limits_{n\,=\,1}^{\infty }{\frac{x^{n}}{n}} \right\}\,dx}=-\int_{0}^{1}{x\ln (1-x)\,dx},$

and we can kill this with standard techniques.