# Convergence and divergence help

1. Dec 9, 2008

### kidsmoker

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

Determine whether the following converges or diverges:

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

with the sum going from n=1 to n=infinity.

2. Relevant equations

Comparison Test, Ratio Test.

3. The attempt at a solution

This should be do-able with the above two tests, since those are the only ones covered in the book up to this point. I think it diverges but I can't seem to get anywhere with either test. Any help appreciated!

2. Dec 9, 2008

### randomlogic78

Re: Convergence/divergence

I think you are going to want to use the ratio test... try setting it up this way...

$$\lim_{n\rightarrow\infty} \frac{\frac{(n+1)^{n+1}}{((n+1)+1)^{((n+1)+1)}}}{\frac{n^n}{(n+1)^{(n+1)}}}$$

3. Dec 9, 2008

### Dick

Re: Convergence/divergence

You can try that. But it's going to give you a limit of 1. So the test is inconclusive. What you really need to do is find a way to use the famous limit (1+1/n)^n=e.

4. Dec 9, 2008

### gabbagabbahey

Re: Convergence/divergence

The ratio test is inconclusive (check for yourself using randomlogic's setup); why don't you show us what you've tried for the comparison test?

5. Dec 9, 2008

### randomlogic78

Re: Convergence/divergence

Indeed it is inconclusive.

6. Dec 9, 2008

### Dick

Re: Convergence/divergence

Reasonable suggestion, nonetheless.

7. Dec 10, 2008

### kidsmoker

Re: Convergence/divergence

Okay thanks, I think i got it:

$$\frac{n^{n}}{(n+1)^{n+1}} = \frac{1}{(n+1)}(\frac{n}{n+1})^{n} = \frac{1}{(n+1)}\frac{1}{(\frac{n+1}{n})^{n}} = \frac{1}{(n+1)}\frac{1}{(1+\frac{1}{n})^{n}}} \geq \frac{1}{(n+1)}\frac{1}{(e+1)}$$

using the fact that (1+1/n)^n tends to e.

Now $$\frac{1}{e+1}\sum{\frac{1}{n+1}}$$ diverges, therefore our series must diverge too :-)

Does that seem okay?

8. Dec 10, 2008

### Dick

Re: Convergence/divergence

That's it. (1+1/n)^n is actually <=e, isn't it? But there's nothing wrong with playing it safe and saying <=(e+1) either.