1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "$\sum e^{a_n}$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It converges absolutely because $a_n^2e^{a_n}\ge0$ :-) $\endgroup$ – robjohn Dec 15 '18 at 8:50
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.