Does this converge?

You mean as n goes to infinity
[tex]ne^{-n^2}[/tex] goes to infinity?
What happens to [tex]e^{-n}[/tex] as n goes to infinity?

 

I see now that it's a series. Use the integral test to show it Converges.
CC

EDIT:Your second question:
1+1/8+2/27+2/64+1/125 +.....can't go to zero because 1+(a bunch of positive things) can't be zero.

 

No, saying it appears to do anything is not sufficient (nor "safe"). I agree with happyg1: use the integral test. This series will converge as long as the corresponding integral
[tex]\int_0^\infty xe^{-x^2}dx[/tex]
converges and that is an easy integral.

Notice that happyg1 gave the obvious answer: 1+ positive numbers can't converge to 0 because its partial sums are always larger than 1! You may be confusing this with the limit of the underlying sequence. If the sequence does not converge to 0, the sum cannot converge but here that sequence does converge to 0 which doesn't tell you anything.

Do you recognize that your sum is
[tex]\sum_1^\infty \frac{1}{n^3}[/itex]

Do you know the "p-series" criterion? For what values of p does
[tex]\sum_1^\infty \frac{1}{n^p}[/itex]
converge?

The integral test would also work nicely for this. Does
[tex]\int_1^\infty \frac{1}{x^3} dx[/tex]
converge?