# Convergence of $\sum_{n=1}^\infty {a_n}^2e^{a_n}$

I'm having a problem solving this question: Suppose the series $\sum_{n=1}^\infty a_n$ converges absolutely. Determine if the following series converges $\sum_{n=1}^\infty {a_n}^2e^{a_n}$

I'm not really sure how to solve this. It looks like it should converge... because $\sum_{n=1}^\infty {a_n}^2$ should converge and $\sum_{n=1}^\infty e^{a_n}$ also looks like it should converge... and I'm not sure but I think multiplications of convergent series results in a convergent series.

Could someone please give me a hand? I would be grateful.

• "$\sum e^{a_n}$ also looks like it should converge". No this series is very divergent: the general term approaches $1$ as $n\to \infty$. – Winther Dec 15 '18 at 11:07

That means $|a_n|$ is bounded since it converges to $0$, thus for $n > N_0, |a_n| < M$, hence:

$|a_n^2\cdot e^{a_n}| < Me^{M}|a_n|$ , for $n > N_0$, and by comparison test, the latter series converges absolutely as well.

• It converges absolutely because $a_n^2e^{a_n}\ge0$ :-) – robjohn Dec 15 '18 at 8:50

let $$b_n = a_n{^2}$$ and $$c_n = a_n^2 e^{a_n}$$

Convergence of series $$a_n$$ => Convergence of series $$a_n{^2}$$

apply limit comparison test(LCT) on $$b_n$$ and $$c_n$$.

$$r = \lim c_n / b_n = \lim e^{a_n}$$

[ By nth term test, $$\lim a_n = 0$$]

therefore $$r=\lim e^0 = 1(\ne0)$$

since summation $$b_n$$converges, by LCT summation $$c_n$$ converges

• Be careful: absolute convergence of series $a_n$ implies the convergence of series $a_n^2$. – robjohn Dec 15 '18 at 8:48