Summation Help

**Peleus** · August 23rd 2009, 05:25 PM

Hi all,

Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

$ \sum (\frac{1}{k^2} - \frac{1}{(k-1)^2}) $

The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

Any help is appreciated.

**Krizalid** · August 23rd 2009, 05:36 PM

That's actually a telescoping sum. Do you know what a telescoping sum is?

**Peleus** · August 23rd 2009, 05:37 PM

Nope, any chance you can point me in the right direction?

I'm guessing it has to do with the fact it's 1/16 - 1/9, then 1/25 - 1/16 etc etc, with the numbers shifting, but I'm not sure of the best way to deal with it.

**Krizalid** · August 23rd 2009, 05:39 PM

Read this: Telescoping Sum -- from Wolfram MathWorld

**skeeter** · August 23rd 2009, 05:40 PM

Originally Posted by Peleus

Hi all,

Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

$ \sum (\frac{1}{k^2} - \frac{1}{(k-1)^2}) $

The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

Any help is appreciated.

$\sum_{k=4}^{24} \frac{1}{k^2} = \frac{1}{4^2} + \frac{1}{5^2} + ... + \frac{1}{24^2}$

$\sum_{k=4}^{24} -\frac{1}{(k-1)^2} = -\frac{1}{3^2} - \frac{1}{4^2} - ... - \frac{1}{23^2} $

sum these two series and what two terms remain?

**Peleus** · August 23rd 2009, 06:33 PM

Thanks guys,

So essentially I'm left with

$ \frac{1}{24^2} - \frac{1}{3^2} $

For the answer.

Appreciate it folks.

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