using the identity to evaluate a sum

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September 15th 2010, 08:24 PM
alice8675309

using the identity to evaluate a sum

My question states: Use the identity 1/( $k^2$ -1)= 1/2(1/(k-1)-1/(k+1)) to evaluate $Sigma$ k=1 to n 1/( $k^2$ -1)

so I have 1/2( $\Sigma$ k=1 to n 1/(k-1)-1/(k+1)). I feel like from here i'm supposed to manipulate the 1/k-1 to some how get it so I can have a_j =1/(j+1) and then replace all the 1/k+1 by a_j..however i'm not exactly sure how to go about this. Would it turn into $\Sigma$ 1/a_j-2 -a_j ?? if so what do i do from here?
September 15th 2010, 09:18 PM
Prove It

First of all, your sum can't possibly start with $k = 1$ because it gives a $0$ denominator, so I assume you mean for it to start at $k = 2$ .
Second, it really helps to use LaTeX so that everyone can read what you are writing.

You are asked to evalute $\sum_{k = 2}^{n}\frac{1}{k^2 - 1}$ .

You know that $\frac{1}{k^2 - 1} = \frac{1}{2}\left(\frac{1}{k - 1} - \frac{1}{k + 1}\right)$

so that means $\sum_{k = 1}^{n}\frac{1}{k^2 - 1} = \frac{1}{2}\sum_{k = 2}^n\left(\frac{1}{k - 1} - \frac{1}{k+ 1}\right)$

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

Everything but the first and last two terms cancel, which leaves

$= \frac{1}{2}\left[1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n + 1}\right]$

$= \frac{1}{2}\left[\frac{3}{2} - \frac{1}{n} - \frac{1}{n + 1}\right]$

$= \frac{1}{2}\left[\frac{3n(n + 1) - 2(n + 1) - 2n}{2n(n + 1)}\right]$

$= \frac{1}{2}\left[\frac{3n^2 + 3n - 2n -2 - 2n}{2n(n + 1)}\right]$

$= \frac{1}{2}\left[\frac{3n^2 - n - 2}{2n(n + 1)}\right]$

$= \frac{3n^2 - n - 2}{4n(n + 1)}$

$= \frac{(n - 1)(3n + 2)}{4n(n + 1)}$ .
September 15th 2010, 09:35 PM
alice8675309

Hey thanks so much! It was soooo much easier to see when you expand everything out..i don't know why i didn't even bother to expand it but thanks. Oh and i'm sorry for my lack of latex skill, i am trying to use them and i keep referring to a latex table but i seem to be horrible at applying them correctly. I need more practice with it and i will def work on it! i'm sorry it was so hard to read but thank you for clearing it up for everyone to see. (and also i did mean k=2 thats what i have on my scrap paper my bad). Thanks again.