# summation problem

• Jun 8th 2009, 06:47 PM
geoffl
summation problem
I'm trying to figure out this problem and have no clue how to do it. The problem reads:

Use the identity 1/(k^2-1) = (1/2)(1/(k-1) - 1/(k+1)) to evaluate sum k=2 to n 1/(k^2-1)
• Jun 8th 2009, 06:57 PM
Krizalid
the problem is almost solved, now, we have $\displaystyle \frac{1}{k-1}-\frac{1}{k+1}=\left( \frac{1}{k-1}-\frac{1}{k} \right)+\left( \frac{1}{k}-\frac{1}{k+1} \right).$ Then, if you consider $\displaystyle \sum\limits_{k=2}^{n}{\left( \frac{1}{k-1}-\frac{1}{k} \right)},$ does it telescope? what 'bout the other one?
• Jun 8th 2009, 07:00 PM
geoffl
what do you mean "does it telescope"? sorry, im new at this stuff.
• Jun 8th 2009, 08:00 PM
mr fantastic
Quote:

Originally Posted by geoffl
what do you mean "does it telescope"? sorry, im new at this stuff.

Write out the first few terms of the series posted by Krizalid. What do you notice ....?
• Jun 8th 2009, 08:02 PM
VonNemo19
Quote:

Originally Posted by geoffl
what do you mean "does it telescope"? sorry, im new at this stuff.

What he means is that a telescoping sum is a sum that can be simplified by the cancelleling of terms through the addition that it implies.

e.g.

$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})$

If you look carefully at this sum you can see that no matter how large n is that there will only be two terms left standing after what I call the addition wars. check it out

$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})$

$\displaystyle =(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+(a_{n-2}-a_{n-3})+...+(a_{3}-a_{2})+(a_{2}-a_{1})$

So after an uncountable number of n terms, who comes out of this war alive?

$\displaystyle \sum_{i=1}^n(a_n-a_{n-1})=a_{n}-a_{1}$

So now look at your problem with a trained eye. You have the ability to predict who's side you shoulb be on!(Bow)
• Jun 8th 2009, 08:25 PM
geoffl
ahhh, ok. that makes sense.

thanks a lot!