# Series correction

• May 20th 2008, 06:15 AM
simplysparklers
Series correction
Hiya guys!
Could someone correct this for me please?Not sure if I went about it right at all! (Wondering)

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??

Jo (Flower)
• May 20th 2008, 06:23 AM
Isomorphism
Quote:

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 (Wink)

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:

$\frac12\left((1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{(n+2)})\right)$
• May 20th 2008, 06:40 AM
simplysparklers
Thanks Isomorphism! (Bear)
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 (Flower)
• May 20th 2008, 07:07 AM
Isomorphism
Quote:

Originally Posted by simplysparklers
Hiya guys!
Could someone correct this for me please?Not sure if I went about it right at all! (Wondering)

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.
• May 20th 2008, 07:28 AM
simplysparklers
(Bear) Hugs for Isomorphism!!! :D Thank you!! (Star)
• May 20th 2008, 05:43 PM
Krizalid
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.