# Is this series convergent or divergent.

1. Dec 14, 2013

### nothingkwt

1. The problem statement, all variables and given/known data

Ʃ ne(-n2)

2. Relevant equations

3. The attempt at a solution

I used the ratio test and wanted to know if the way I did it is correct or not

|a(n+1) / a(n)|

n+1 (e(-n2 -2n-1)) / n e(-n2)

Now e-n^2 cancels and we get

limn→∞ n+1/n * 1/(e2n)(e)

After you take the limits you get (1)*0 = 0 < 1 so it's convergent

2. Dec 14, 2013

### pasmith

3. Dec 14, 2013

### Curious3141

Sure, this works. But this sum looks almost purpose-built for the integral test. Try it (if you're allowed to use it).

4. Dec 14, 2013

### nothingkwt

I am actually but I just wanted to see if the ratio test worked

Thanks for the replies!

5. Dec 14, 2013

### Ray Vickson

You really need to learn how to write in ASCII---in particular, you need to use brackets. Your first equation AS WRITTEN says
$$\left|\frac{a_{n+1}}{a_n}\right| = n +1 \frac{e^{-n^2-2n-1}}{n} e^{-n^2}.$$
$$\lim_{n \to \infty} n + \frac{1}{n} \frac{1}{e^{2n}} e$$
I hope these are not what you mean. I hope you intended the first one to be
$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)e^{-n^2-2n-1}}{n e^{-n^2}},$$
etc. To make sure this happens you need parentheses!

6. Dec 15, 2013

### nothingkwt

Yeah I'm not very good with ASCII I'm still learning how to use them.