# Calculating series sum

• Dec 7th 2008, 12:30 AM
Twig
Calculating series sum
Hi!

This one is giving me problems.

$\sum_{k=1}^{\infty}\, \frac{1}{(2k-1)(2k+1)} = \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5}\, ...$

The book tells me to calculate the n:th partialsum by dividing $a_{k}$ into two or more terms. Then to prove that the series converge and calculate its sum.

What is it that I am supposed to do really, donīt really get it.

Thanks
• Dec 7th 2008, 01:00 AM
mr fantastic
Quote:

Originally Posted by Twig
Hi!

This one is giving me problems.

$\sum_{k=1}^{\infty}\, \frac{1}{(2k-1)(2k+1)} = \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5}\, ...$

The book tells me to calculate the n:th partialsum by dividing $a_{k}$ into two or more terms. Then to prove that the series converge and calculate its sum.

What is it that I am supposed to do really, donīt really get it.

Thanks

Partial fraction decomposition of $a_k$:

$a_k = \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1}\right)$.

You should see what happens next ....
• Dec 7th 2008, 03:45 AM
Krizalid
$\sum\limits_{k=0}^{\infty }{\frac{1}{(2k+1)(2k+3)}}=\frac{1}{3}+\sum\limits_ {k=1}^{\infty }{\frac{1}{(2k+1)(2k+3)}},$ and from there the convergence is easy, since for each $k\ge1$ we have,

$\frac{1}{(2k+1)(2k+3)}=\frac{1}{4k^{2}+8k+3}\le \frac{1}{4k^{2}},$ then your series converges by comparison with $\sum\limits_{k=1}^{\infty }{\frac{1}{k^{2}}}$ which is a $p-$series for $p=2>1.$
• Dec 7th 2008, 05:33 AM
Twig
hi
Hi

Thanks for your help guys, it became a bit clearer for me.
I guess the 'trick' here is to write out some terms when $a_{k}$
has been partial divided, and see what terms cancel out each other?

I got:

$\frac{1}{2}(\frac{1}{2k - 1} - \frac{1}{2k + 1}) + \frac{1}{2}(\frac{1}{2k + 1} - \frac{1}{2k+3})\, ...$

I can see that some terms cancel each other.
So I guess Ill end up with:

$\frac{1}{2}(1 - \frac{1}{2n + 1}) \rightarrow \frac{1}{2}\; \mbox{when}\; n\rightarrow \infty$

Also thanks Krizalid for helping me with the convergence proof.
One question however, why did you write

$\frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)}$ ?

Couldnīt we just make proof from scratch, without explicitly writing out the first term $\frac{1}{3}$ ?
• Dec 7th 2008, 07:58 AM
Krizalid
Quote:

Originally Posted by Twig

One question however, why did you write

$\frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)}$ ?

'cause $\sum\limits_{k=1}^{\infty }{\frac{1}{(2k-1)(2k+1)}}=\sum\limits_{k=0}^{\infty }{\frac{1}{(2k+2-1)(2k+2+1)}}=\sum\limits_{k=0}^{\infty }{\frac{1}{(2k+1)(2k+3)}}.$

Now I just considered the last sum and summed the first term to start the sum in $k=1.$
• Dec 7th 2008, 02:20 PM
mr fantastic
Quote:

Originally Posted by Twig
Hi

Thanks for your help guys, it became a bit clearer for me.
I guess the 'trick' here is to write out some terms when $a_{k}$
has been partial divided, and see what terms cancel out each other?

I got:

$\frac{1}{2}(\frac{1}{2k - 1} - \frac{1}{2k + 1}) + \frac{1}{2}(\frac{1}{2k + 1} - \frac{1}{2k+3})\, ...$

I can see that some terms cancel each other.
So I guess Ill end up with:

$\frac{1}{2}(1 - \frac{1}{2n + 1}) \rightarrow \frac{1}{2}\; \mbox{when}\; n\rightarrow \infty$

Also thanks Krizalid for helping me with the convergence proof.
One question however, why did you write

$\frac{1}{3} + \sum_{k=1}^{\infty}\, \frac{1}{(2k+1)(2k+3)}$ ?

Couldnīt we just make proof from scratch, without explicitly writing out the first term $\frac{1}{3}$ ?

As an alternative:

$\frac{1}{(2k - 1)(2k + 1)} = \frac{1}{4k^2 - 1} \leq \frac{1}{4k^2 - k^2} = \frac{1}{3k^2} = \frac{1}{3} \cdot \frac{1}{k^2}$

and convergence is assured by the comparison test.