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)

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- Jun 8th 2009, 06:47 PMgeofflsummation 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 PMKrizalid
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 PMgeoffl
what do you mean "does it telescope"? sorry, im new at this stuff.

- Jun 8th 2009, 08:00 PMmr fantastic
- Jun 8th 2009, 08:02 PMVonNemo19
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 PMgeoffl
ahhh, ok. that makes sense.

thanks a lot!