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.

  • $\begingroup$ "$\sum e^{a_n}$ also looks like it should converge". No this series is very divergent: the general term approaches $1$ as $n\to \infty$. $\endgroup$ – 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.

  • $\begingroup$ It converges absolutely because $a_n^2e^{a_n}\ge0$ :-) $\endgroup$ – 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

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

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