# Infinite Series Problem

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• Oct 8th 2010, 07:03 PM
joeseinfeld
Infinite Series Problem
Having trouble with this problem, infact this whole topic, but thats another story

The Nth partial sum, and the convergence behaviour, of the telescoping series

¥
[SIZE=+ 3]å
n = 1
2 (2n + 1)(2n + 3)
[/SIZE]
how do i go about solving this? any help would be appreciated,

thanks
• Oct 8th 2010, 07:29 PM
Chris L T521
Quote:

Originally Posted by joeseinfeld
Having trouble with this problem, infact this whole topic, but thats another story

The Nth partial sum, and the convergence behaviour, of the telescoping series

¥
[SIZE=+ 3]å
n = 1
2 (2n + 1)(2n + 3)
[/SIZE]
how do i go about solving this? any help would be appreciated,

thanks

I'm sorry to say, but I think its hard to decipher what the series in question is. That's why we like our members to learn some basic $\displaystyle \text{\LaTeX}$ (see the attachment to this post)

By any chance, could it possibly be $\displaystyle \displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2( 2n+1)(2n+3)}$ or $\displaystyle \displaystyle\sum\limits_{n=1}^{\infty}\frac{2}{(2 n+1)(2n+3)}$?

Please inform me which one you mean, and I'll fix it in your original post.
• Oct 10th 2010, 05:33 PM
joeseinfeld
Quote:

Originally Posted by Chris L T521
I'm sorry to say, but I think its hard to decipher what the series in question is. That's why we like our members to learn some basic $\displaystyle \text{\LaTeX}$ (see the attachment to this post)

By any chance, could it possibly be $\displaystyle \displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2( 2n+1)(2n+3)}$ or $\displaystyle \displaystyle\sum\limits_{n=1}^{\infty}\frac{2}{(2 n+1)(2n+3)}$?

Please inform me which one you mean, and I'll fix it in your original post.

Yes it is the second one with the 2 ontop. sorry i didn't notice it came out like that.

Thanks again
• Oct 10th 2010, 07:51 PM
TheCoffeeMachine
Quote:

Originally Posted by joeseinfeld
Yes it is the second one with the 2 ontop.

Hint: $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}$.
• Oct 10th 2010, 08:21 PM
Chris L T521
Quote:

Originally Posted by TheCoffeeMachine
Hint: $\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}$.

I would actually consider a finite series first and focus on the expansion of $\displaystyle \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)$ because its easier to see the numerous cancellations. Then, the OP should take $\displaystyle k\to\infty$ in that result to complete the problem.
• Oct 10th 2010, 08:40 PM
TheCoffeeMachine
Quote:

Originally Posted by Chris L T521
I would actually consider a finite series first and focus on the expansion of $\displaystyle \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)$ because its easier to see the numerous cancellations.

I don't really see how the cancellations are easier to see in the finite case.
If anything, that seems unnecessary to me. But perhaps I'm missing something.
Spoiler:
$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1} = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{1}{2n+3}$
• Oct 11th 2010, 02:24 AM
Unbeatable0
Quote:

Originally Posted by TheCoffeeMachine
I don't really see how the cancellations are easier to see in the finite case.
If anything, that seems unnecessary to me. But perhaps I'm missing something.
Spoiler:
$\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1} = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{1}{2n+3}$

These series are divergent to infinity.
• Oct 11th 2010, 03:37 AM
HallsofIvy
The series TheCoffeeMachine shows in his post are divergent but that is not the original problem. What do you get if you subtract $\displaystyle \sum_{n=1}^\infty \frac{1}{2n+3}$ from that?

Chris L T521's point with "consider finite series first" was that if n= 1, the sum $\displaystyle \frac{1}{3}- \frac{1}{5}$. If n= 2, $\displaystyle \frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}= \frac{1}{3}- \frac{1}{7}$, if n= 3, $\displaystyle \frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}+ \frac{1}{7}- \frac{1}{9}= \frac{1}{3}- \frac{1}{9}$, etc.
• Oct 11th 2010, 10:54 AM
Krizalid
Quote:

Originally Posted by Chris L T521
I would actually consider a finite series first and focus on the expansion of $\displaystyle \displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)$ because its easier to see the numerous cancellations.

instead of doing that, having $\displaystyle \dfrac1{2n+1}-\dfrac1{2n+3},$ consider the first term, if we make $\displaystyle n\to n+1$ that yields the second term, hence, telescoping series and that's all.
• Oct 12th 2010, 01:59 AM
joeseinfeld
thanks guys figured it out after reading your replies

i got Sn=(1/3)-1(2N+3).