Series Comparison Test for (n^n)/n! and Convergence Analysis

[phrygian]

Homework Statement

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

Homework Equations

p = an+1/an

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, don't (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!

 

[Bohrok]

Why not try the ratio test?

 

[Office_Shredder]

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?

 

[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?

 

[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