Solve Simple Series Sum: Find Sum or DNE

[tangibleLime]

Homework Statement

Find the sum of the series, if it converges. Otherwise, enter DNE.

\Sigma = \frac{7}{n(n+2)}

The sigma is from n=2 to infinity.

Homework Equations

The Attempt at a Solution

I really don't know how to go about this problem. My professor has tenure and he's pretty much checked out at this point, and the book is a $100 paper weight.

I tried to get it to look like a geometric series by factoring the bottom into n^2+2n, pulling out the 7 and ending up with
7 \Sigma \frac{1}{n^{2}+2n}

But this looks wrong and I have no idea what to do. A push in the right direction would be great.

 

[Char. Limit]

Well, first thing I'd do is check that it DOES converge. You can worry about finding a sum later. As for finding if it converges, either an integral test or a comparison test to 1/n^2 would work.

 

[dickdick]

Homework Statement

Find the sum of the series, if it converges. Otherwise, enter DNE.

\Sigma = \frac{7}{n(n+2)}

The sigma is from n=2 to infinity.

Homework Equations

The Attempt at a Solution

I really don't know how to go about this problem. My professor has tenure and he's pretty much checked out at this point, and the book is a $100 paper weight.

I tried to get it to look like a geometric series by factoring the bottom into n^2+2n, pulling out the 7 and ending up with
7 \Sigma \frac{1}{n^{2}+2n}

But this looks wrong and I have no idea what to do. A push in the right direction would be great.

It's not geometric. Compare with a power series. What do you know about series of the form 1/n^p? Does the $100 paperweight have anything to say about them?

 

[ns2675]

I'm going to assume that you have found out how to prove convergence. I'm also going to divide by 7.

To evaluate this sum, you're going to want to search for a pattern. Start by looking at the terms corresponding to n, n+2, n+4, ... . For consecutive terms of this form we have

1/n(n+2) + 1/(n+2)(n+4) = 2/n(n+4).

See if you can find the pattern; then apply your findings individually to the the terms for which n is even and those for which n is odd.

 

[The legend]

take 7 out of the sigma and write
1/n(n+2) = [1/n - 1/(n+2)]/2

Take 1/2 out too.

now find term 1, term 2, term 3 and you will see a pattern where many things get cancelled. Add the rest, and multiply by the 7/2.PS: I assume you know the convergence proof.