Is the series (n!)/(n^n) convergent or divergent?

[joe_cool2]

I am to find whether the sum of (n!)/(n^n) converges or diverges. I tried both the limit comparison test, and a regular comparison test. (These are the only types of tests I am allowed to use.) So I tried several approaches:

Approach #1: (n!)/(n^n) > 1/(n^n)

Normally we use a setup like this to prove something with a p-series. However, the expression on the left side of the inequality isn't a p-series.

Approach #2: (n!)/(n^n) < n!

While this expression is true, it is not useful because the formula for the series is less than, not greater than, the series that is known to diverge.

Approach #3 (Limit comparison): a_n = (n!)/(n^n) ; b_n = 1/(n^n)

a_n/b_n = n!

The limit here is, unfortunately, infinite, and I have to stop here.

What other approach can I take that would result in more success?

 

[chiro]

Hey joe_cool2 and welcome to the forums.

My intuitive guess is that your function will converge. One test to show this that seems appropriate would be the ratio test. Take a look at this page:

http://en.wikipedia.org/wiki/Ratio_test

 

[joe_cool2]

Hello, thanks for the welcome. It is much appreciated.

I am perfectly aware that it is often wise to use the ratio test in these situations with n!. However, I have been restrained to using specifically those two techniques mentioned earlier. Any ideas?

 

[micromass]

The idea is to compare your function to a geometric series. That is, you have to find an $0\leq a<1$ such that

$$\frac{n!}{n^n}\leq a^n$$

for all n. If you can find such an a, then your series will converge. So we need to find an a such that

$$\frac{n!}{(an)^n}<1$$

Try to show that the left hand side is decreasing from a certain point on.

 

[DryRun]

Whenever factorial is involved in series, the ratio test is your best bet.

 

[micromass]

Whenever factorial is involved in series, the ratio test is your best bet.

He's not allowed to use it. Please read the thread first before responding.

 

[Bohrok]

Can you show that $\sum\frac{n^n}{(n!)^n}$ converges? If you can, then use the limit comparison test with that and your series.