# Determine whethere the following series is convergent or divergent

1. Jan 24, 2010

### Ratio Test =)

1. The problem statement, all variables and given/known data
Determine whether the following series is convergent or divergent:
$$\frac{1}{2^2}+\frac{2^2}{3^3}+\frac{3^3}{4^4}+....$$
I rewrite it as:
$$\sum_{n=1}^{\infty} \frac{n^n}{(n+1)^{n+1}}$$

2. Relevant equations

3. The attempt at a solution
I stopped.
I can not do anything.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 24, 2010

### jgens

You do need to show us your attempt at the solution before we can help, so which convergence tests have you tried?

3. Jan 24, 2010

### Ratio Test =)

Its not geometric
not telescoping
nth term test failed
can not use integral test
Ratio & Root failed because its algebric function

There is only the comparison tests
I tried them a lot
but every time i failed :/

4. Jan 24, 2010

### Dick

Write it as (n^n/(n+1)^n)*(1/(n+1)). The first factor approaches a finite limit. What is it?

5. Jan 24, 2010

### Ratio Test =)

The first one goes to 1/e
the second goes to 0
1/e times 0 = 0

nth term test failed.

6. Jan 24, 2010

### jgens

Alright, I would suggest using the comparison test for this particular series (although I'm sure that there's probably a simpler way to do it). If $n \geq 1$ can you find an upper bound for the following:

$$\frac{(n + 1)^{n+1}}{n^{n+1}}$$

Edit: Fixed typo

Edit 2: Fixed another typo. Sorry, I just seem to be mistake prone today.

Last edited: Jan 24, 2010
7. Jan 24, 2010

### Dick

What 'nth term test' are you talking about? You are right the first factor approaches 1/e. So for large values of n a term in the series is approximately 1/(e*(n+1)). What kind of test does that suggest?

8. Jan 24, 2010

### Ratio Test =)

nth terms test = Test for divergence.

hmmm .. I got it .. Limit comparison test with the divergent series $$\sum_{n=1}^{\infty} \frac{1}{n+1}$$
The limit of the limit comparison test will give 1/e which is positive and finite number
hence, our series diverges.

Right?

Last edited: Jan 24, 2010
9. Jan 24, 2010

Right.