However, it apparently converges according to the answer in the back of my text (I guess $\displaystyle e^{n^2}$ end up growing much faster than $\displaystyle n!$ which is kind of surprising to me because obviously $\displaystyle n!$ grows much faster than just $\displaystyle e^n$ for even relatively small n). In any case, I'd appreciate it if someone can show me the correct solution.

In calculus II the ratio test is the only test we have for convergence of a series involving factorials. I suspect my problem was purely algebraic.

Thanks ~

By the way, when people write $\displaystyle e^{(n+1)^2}$ do they mean distribute the 2 across the originally raised expression (as would normally be the case following the law of exponents) or do they mean to actually square the originally raised expression?

In other words, is the above expression equivalent to

1.) $\displaystyle e^{(n+1)^2} = e^{2n+2} = e^2n\cdot e^2 $

1.) $\displaystyle e^{(n+1)^2} = e^{n^2+2n+1} = e^{n^2}\cdot e^{2n}\cdot e^1 = e^2n\cdot e^2n\cdot e = e^{4n+1}$

I'm pretty sure it has to be the first.....