1. ## 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

2. 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 $\text{\LaTeX}$ (see the attachment to this post)

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

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

3. 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 $\text{\LaTeX}$ (see the attachment to this post)

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

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

4. Originally Posted by joeseinfeld
Yes it is the second one with the 2 ontop.
Hint: $\displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}$.

5. Originally Posted by TheCoffeeMachine
Hint: $\displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)} = \sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right) = \sum_{n=1}^{\infty}\frac{1}{2n+1}-\sum_{n=1}^{\infty}\frac{1}{2n+3}$.
I would actually consider a finite series first and focus on the expansion of $\displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)$ because its easier to see the numerous cancellations. Then, the OP should take $k\to\infty$ in that result to complete the problem.

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

7. 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:
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1} = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{1}{2n+3}$
These series are divergent to infinity.

8. The series TheCoffeeMachine shows in his post are divergent but that is not the original problem. What do you get if you subtract $\sum_{n=1}^\infty \frac{1}{2n+3}$ from that?

Chris L T521's point with "consider finite series first" was that if n= 1, the sum $\frac{1}{3}- \frac{1}{5}$. If n= 2, $\frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}= \frac{1}{3}- \frac{1}{7}$, if n= 3, $\frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}+ \frac{1}{7}- \frac{1}{9}= \frac{1}{3}- \frac{1}{9}$, etc.

9. Originally Posted by Chris L T521
I would actually consider a finite series first and focus on the expansion of $\displaystyle\sum_{n=1}^k\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)$ because its easier to see the numerous cancellations.
instead of doing that, having $\dfrac1{2n+1}-\dfrac1{2n+3},$ consider the first term, if we make $n\to n+1$ that yields the second term, hence, telescoping series and that's all.