# Does this series converege?

1. Feb 10, 2008

### Mattofix

1. The problem statement, all variables and given/known data

The series is shown in the answer

3. The attempt at a solution

2. Feb 10, 2008

### Mattofix

i feel like im going in the totally wrong direction - if so please can someone please point me the right way.

3. Feb 10, 2008

### Dick

Expand the denominator first. The n^2 terms cancel so the denominator is basically proportional to n. The numerator, I think, tends to 1. Do you agree? So do you think it converges or diverges? It's good to have opinion to organize the strategy before you start trying to prove anything.

4. Feb 10, 2008

### rootX

I am getting:

1+exp(-n)
----------
4n.e^n

Now, using ratio test I can prove that 1/(4n.e^n) is converging, and I also know that 1+exp(-n) is converging so that means there product is converging?

My Second approach

lim n--> inf (this series)/ [(1+exp(n))/exp(n)] is equal to 0, so this is equivalent to:
exp(n)+1
----------
exp(n)

now this seems wrong because my series does not converge. What have done wrong here?

5. Feb 11, 2008

### Dick

The nth term in the series is (1+e^(-n))/((n+1)^2-(n-1)^2). The denominator is 4n. The numerator converges to one. It's begging for a comparison test.

6. Feb 11, 2008

### Mattofix

thanks.

(1+exp(-n))/4n > 1/4n

and since 1/4n diverges so must the series.