# Need to evaluate an inf. series

1. Sep 19, 2006

### ToxicBug

For a series that is Sigma (n!)/(n^n), I did the ratio test and now I need to prove that limit (n^n)/(n+1)^n < 1, can someone give me a hand on how I can do this?

2. Sep 19, 2006

### shmoe

hint:the limit of $$\left(\frac{n}{n+1}\right)^{-n}$$ should be very familiar...

3. Sep 20, 2006

### ToxicBug

Never seen it before...

4. Sep 20, 2006

### CRGreathouse

Caluclate its numeric value for n=1, n=2, n=10, and n=1000.

5. Sep 20, 2006

### shmoe

You've never seen:

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

before?

Okie, try a comparison of your original sum with $$\sum_{n=1}^\infty\frac{1}{2n^2}$$

6. Sep 20, 2006

### ToxicBug

1/e, you could've just said it... as for your second suggestion, how the hell are the two series similar? Are you joking?

7. Sep 21, 2006

### shmoe

I could have also done the whole problem for you, but what's the point of that? I make no apologies for expecting you to meet me well over half way.

No joke. Though I did mean the 2 to be in the numerator to make it more obvious (it doesn't end up mattering though, the terms in your series decay very fast), namely $$\sum_{n=1}^\infty\frac{2}{n^2}$$.

Break the terms of your series up into 'n' factors:

$$\frac{n!}{n^n}=\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\ldots\left(\frac{n-1}{n}\right)\left(\frac{n}{n}\right)$$

Now compare with $$\frac{2}{n^2}$$.