Looking for the limit if it exists

[Sheneron]

Homework Statement

$$\sum_{n=2}^{\infty} \frac{2}{n^2-1}$$

The Attempt at a Solution

I don't know what it is about this, maybe just because its late, but I can't figure this out. Can someone please help. Just looking for the limit if it exists. I have tried to figure it out; just not sure what to do. I wrote down the the terms of the sequence and the sum, but I can't see a pattern.

 

[dirk_mec1]

Note that: $$\sum_{k=1}^{\infty} \frac{1}{k^2}= \frac{ \pi ^2}{6}$$

 

[weejee]

Note the identity $$\cot x = \sum_{n=-\infty}^{\infty} \frac{1}{x+n \pi} = \frac{1}{x} + \sum_{n=1}^{\infty} \frac{2x}{x^2 - n^2 \pi^2}$$, which can be proved by differentiating ln(sin(x)) using the infinite product form of sin(x).

Then, using coth(x) = i cot(ix) we get $$\coth x = \frac{1}{x} + \sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2 \pi^2}$$, which may be useful for your problem.

 

[weejee]

Oops.. you can use the integral test if you just want to check its convergence.

 

[Sheneron]

Thanks I figured this out today though, and just for completeness I will post the way I solved it.

If you use partial fraction decomposition to split of the initial fraction it becomes a telescoping series.

$$\frac{2}{n^2-1} = \frac{2}{(n+1)(n-1)}$$
$$\frac{2}{(n+1)(n-1)} = \frac{A}{n+1} + \frac{B}{n-1}$$

$$2 = A(n-1) + B(n+1)$$

Solve that for A and B and you get A = -1, B=1.

So the original sum becomes:

$$\sum_{n=2}^{\infty} \frac{1}{n-1} - \frac{1}{n+1}$$

if you write plug in some terms and write out the series you will get:

$$S_{n} = 1 + \frac{1}{2} -\frac{1}{n} -\frac{1}{n+1}$$


$$\lim_{x -> \infty} S_{n} = 1 + \frac{1}{2} = 3/2$$

 

[dirk_mec1]

Thanks I figured this out today though, and just for

if you write plug in some terms and write out the series you will get:

$$S_{n} = 1 + \frac{1}{2} -\frac{1}{n} -\frac{1}{n+1}$$


$$\lim_{x -> \infty} S_{n} = 1 + \frac{1}{2} = 3/2$$

Really?

I thought of something like:

S_n= 1-1/2+1/2-1/3 +1/3-1/4 +1/4-1/5+...=1

\\edit: your sum in the first post is different than in the fifth!

 

[Office_Shredder]

Except that's not the telescoping sum that he found. So you would get

1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6

Confusion abounds however, is the series in the original post not the one you were trying to sum?

 

[dirk_mec1]

Except that's not the telescoping sum that he found. So you would get

1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + 1/4 - 1/6

Yes, you're right I didn't read well enough!

 

[Sheneron]

Oh yeah sorry, in the very first post it should be a minus sign. Boy I guess that really makes a big difference.