1. ## Telescoping series sum

so i have a telescoping series that converges and I need to find the sum. How do I do this?

Thanks

2. series from zero to infinite of 4/((x squared)-1)

sorry about the way i wrote it out i will learn with time

3. Originally Posted by dogface
series from zero to infinite of 4/((x squared)-1)

sorry about the way i wrote it out i will learn with time
Ok you have $\sum_{n=0}^{\infty}\frac{4}{x^2-1}\Rightarrow\sum_{n=0}^{\infty}\frac{4}{(x-1)(x+1)}$

then by using partial fractions decomposition we get

$\sum_{n=0}^{\infty}\bigg[\frac{2}{x-1}-\frac{2}{x+1}\bigg]$...now write out the first few terms of the sequence...do you notice a pattern of cancellations?

4. got it thank you

5. Originally Posted by dogface
got it thank you
No problem! See we are not so bad here!

6. Originally Posted by dogface
series from zero to infinite of 4/((x squared)-1)

sorry about the way i wrote it out i will learn with time
There is a problem: What happens when x = 1 ......? So assuming you instead want

$= \sum_{x = 2}^{\infty} \left( \frac{2}{x-1} - \frac{2}{x+1} \right) = 2 \sum_{x = 2}^{\infty} \left( \frac{1}{x-1} - \frac{1}{x+1} \right)$

using partial fractions

$= 2 \left( \left[ 1 - \frac{1}{3}\right] + \left[ \frac{1}{2} - \frac{1}{4}\right] + \left[ \frac{1}{3} - \frac{1}{5}\right] + \left[ \frac{1}{4} - \frac{1}{6}\right] + .....\right)$

and yu should be able to see that after cancelling you're left with $2 \left(1 + \frac{1}{2} \right) = 3$.

7. Originally Posted by Mathstud28
Ok you have $\sum_{n=0}^{\infty}\frac{4}{x^2-1}\Rightarrow\sum_{n=0}^{\infty}\frac{4}{(x-1)(x+1)}$
This is abuse of notation. When something implies another thing, a previous equality symbol should be there, so instead an $\implies$ that's actually an $=.$

As for dogface's question, I think you should find problems about your issue and post it (by makin' new threads of course) to see where you're getting stuck.