# Thread: using the identity to evaluate a sum

1. ## using the identity to evaluate a sum

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

so I have 1/2($\displaystyle \Sigma$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 $\displaystyle \Sigma$ 1/a_j-2 -a_j ?? if so what do i do from here?

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

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

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

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

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

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

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

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

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

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

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

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

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

3. 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.