Ratio test for $Σ ne^{-n^2}$?

Question

How do I show that the infinite series

$\sum_{n=1}^{\infty}{ne^{-n^2}}$

is converging using the ratio test?

I know that the result $p$ has to be less than 1, so I have already used the theorem and gotten:

$p = \lim_{n\to\infty}{\frac{a_{n+1}}{a_n}}$

which gives $\frac{(n+1)e^{-(n+1)^2}}{ne^{-n^2}}$

so what I struggle with is the next step, to simplify it. Is there a rule or way you guys could show me or advise me so I can solve this and similar problems with? the textbook over complicates everything

thankyou so much in advance:)

Answer 1

Note that $$ \frac{(n+1)e^{-(n+1)^2}}{ne^{-n^2}}=\frac{n+1}{n}\cdot e^{-(n+1)^2+n^2}=\frac{n+1}{n}\cdot e^{-2n-1}. $$ The first term converges to $1$ as $n\to\infty$; what does the second term do?

Answer 2

Well, $\frac{n+1}{n}\to1,$ so you only have to deal with $\frac{e^{-(n+1)^2}}{e^{-n^2}},$ and the laws of exponents take care of that.

Answer 3

Others have shown the ratio test. The root test also works.

$(ne^{-n^2})^{1/n} =n^{1/n}e^{-n} $ and this goes to zero quite strongly (all that is needed is that is is less than 1).