so i have a telescoping series that converges and I need to find the sum. How do I do this?
Thanks
Ok you have $\displaystyle \sum_{n=0}^{\infty}\frac{4}{x^2-1}\Rightarrow\sum_{n=0}^{\infty}\frac{4}{(x-1)(x+1)}$
then by using partial fractions decomposition we get
$\displaystyle \sum_{n=0}^{\infty}\bigg[\frac{2}{x-1}-\frac{2}{x+1}\bigg]$...now write out the first few terms of the sequence...do you notice a pattern of cancellations?
There is a problem: What happens when x = 1 ......? So assuming you instead want
$\displaystyle = \sum_{x = 2}^{\infty} \left( \frac{2}{x-1} - \frac{2}{x+1} \right) = 2 \sum_{x = 2}^{\infty} \left( \frac{1}{x-1} - \frac{1}{x+1} \right)$
using partial fractions
$\displaystyle = 2 \left( \left[ 1 - \frac{1}{3}\right] + \left[ \frac{1}{2} - \frac{1}{4}\right] + \left[ \frac{1}{3} - \frac{1}{5}\right] + \left[ \frac{1}{4} - \frac{1}{6}\right] + .....\right)$
and yu should be able to see that after cancelling you're left with $\displaystyle 2 \left(1 + \frac{1}{2} \right) = 3$.
This is abuse of notation. When something implies another thing, a previous equality symbol should be there, so instead an $\displaystyle \implies$ that's actually an $\displaystyle =.$
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.