# Series Comparison Test

1. Oct 21, 2009

### phrygian

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

Find if the sum from n = 1 to infinity of (n^n)/n! diverges or not.

2. Relevant equations

p = an+1/an

3. The attempt at a solution

Using the comparison test I get to the point where p_n = (n+1)^(n+1) / [(n+1) n^n]

Shouldnt p just be 0, dont (n+1)^(n+1) and n^n cancel for large n? The book says the answer is p = e, how do you get there?

Thanks for the help!

2. Oct 21, 2009

### Bohrok

Why not try the ratio test?

3. Oct 21, 2009

### Office_Shredder

Staff Emeritus
nn+1/nn doesn't cancel for large n, so when you replace nn+1 with (n+1)n+1 why would you expect it to?

4. Oct 21, 2009

### Dick

No, they don't cancel for large n. You are jumping to conclusions. If I look at your expression I would write it as (n+1)^n/n^n. Do you see how? Now what?

5. Oct 21, 2009

### lanedance

not zero, cancel the n+1 term
$$\frac{(n+1)^{n+1}}{(n+1) n^n} = \frac{(n+1)^n}{n^n} = (\frac{n+1}{n})^n$$