# Series test ?

1. May 23, 2012

### sid9221

$$\sum_{r=1}^{\infty} \frac{r!}{3^{r^{2}}}$$

My solution:

$$\frac{3^{r^2}}{r!} > r^2$$

So $$\frac{r!}{3^{r^2}} < \frac{1}{r^2}$$

So as 1/r^2 converges, it converges by comparison test.

This was in my exam today, I messed up a lot leading up to it. But the question said I could use any test in general to work this out.

It was work 6 marks which is quite a bit so is my solution okay ?

2. May 23, 2012

### sharks

Use the ratio test. The latter is always preferable whenever factorials are involved.

3. May 23, 2012

### sid9221

It was in my exam, it's already over. I did the ratio test for some reason cut it out and put this instead as it was smaller.

It's correct right ?

4. May 23, 2012

### micromass

Staff Emeritus
How do you know this?? I'm not saying it's wrong, but it should have an explanation.

5. May 23, 2012

### sid9221

I evaluated a few terms during the exams(beginning with 1) its not a little bigger it's ALOOOT bigger. So much so that you can't evaluate beyond r=13 as size of the number is too big.

At r=14 there are more digits in the result than atoms in the universe so it kinda goes without saying that its bigger than r^2.

Regardless, I'm hoping no marks are deducted as we are interested in the asymptotic difference and if it's huge...

Last edited: May 23, 2012
6. May 23, 2012

### sharks

Here's my attempt using ratio test:
$$u_n=\frac{r!}{3^{r^{2}}}$$
$$u_{n+1}=\frac{(r+1)!}{3^{(r+1)^2}}=\frac{(r+1)r!}{3^{r^2+2r+1}}=\frac{(r+1)r!}{3^{r^2}.3^{2r}.3^1}$$
$$L= \lim_{r\to\infty}\frac{u_{n+1}}{u_n}=\lim_{r\to \infty} \left( \frac{(r+1)r!}{3^{r^2}.3^{2r}.3}\times \frac{3^{r^{2}}}{r!} \right) =\lim_{r\to\infty}\frac{(r+1)}{3^{2r}.3}= \lim_{r\to\infty}\frac {(r+1)}{9^r.3}$$The limit is an indeterminate form ∞/∞, so using L'Hopital's rule:$$\lim_{r\to\infty}\frac {1}{9^r.3\ln 9}=\frac{1}{\infty}=0$$
Since L < 1, therefore the original series converges.

Last edited: May 23, 2012