# Proof of convergence and divergence

1. Feb 20, 2015

1. The problem statement, all variables and given/known data
Does \frac{2^{n}}{n!} converge or diverge?

3. The attempt at a solution
Is there more than one way to prove this?
I would appreciate a few directions.
I've been trying the Squeeze theorem for a long time.
I said 1/n! was smaller, but I have no damn idea how to say what's bigger. I COULD figure out how to say \frac{n!}{n^{n}} was convergent, with the squeeze theorem, but this is really difficult.
Is the squeeze theorem the only way?
Would appreciate some good hints.

Wait, what the heck, how do I get latex to work?
Oh well I think you can all figure out what's written
The first equation is plainly (2^n)/n!
The second one is n!/(n^n)

2. Feb 20, 2015

### Ray Vickson

For $f_n = 2^n / n!$, do you really mean you want to know if $t_n$ itself converges (exactly as you have written), or do you actually want to determine if $\sum t_n$ converges? The reason I ask is that often posters write the former when they mean the latter.

3. Feb 20, 2015

Oh, I would like to know if the sequence converges, not the series.
But just wondering, if the sequence does converge it won't mean the series does, right?
Could you help me see if the series converges as well?

4. Feb 21, 2015

### HallsofIvy

Staff Emeritus
One way to look at it is that, for a given n, both n! and $2^n$ have n factors. But n- 2 of the factors of n! are larger than 2.

5. Feb 21, 2015

### Dick

If you've done series, try the ratio test to see if the series converges. If the series converges then the sequence must converge to zero.

6. Feb 21, 2015

### Ray Vickson

You can use $4! > 4 \cdot 3 > 3^2$, $5! > 3^3$, $6! > 3^4$, etc., to get a useful bound on $2^n/n!$. Alternatively, you can use Stirling's Formula for the asymptotic behavior of $n!$, but that seems like overkill in the present case.