# Does the series $\sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$ converge or diverge?

Does the series $$\sum_{n=1}^{\infty}(-1)^{n}n^{2}e^{\frac{-n^{3}}{3}}$$ converge or diverge?

I have attempted using the alternating series test, but couldn't find a useful function to use as $b_n$ and no other tests I know seem to be useful in coming to the conclusion of whether it is convergent or divergent.

• Is the power of the( -1 ) 2? – Elad Apr 18 '17 at 21:08
• oops that is supposed to be n – goldenlinx Apr 18 '17 at 21:10
• So you should show that $f(n)=n^2*e^\frac{-n^3}{3}$ is a deacreasing function for large enough $n$ – Elad Apr 18 '17 at 21:14
• That's not trivial to a student learning it for the first time. :( – tilper Apr 18 '17 at 21:26
• I understand your points but please take mine too into consideration: do we really want MSE to be flooded by Calc-1 almost trivial questions? I am fine with a reasonable amount of them to occupy the main page, but nowadays they are becoming way too many. In my humble opinion of course, and as an active member of MSE I have the power to express it through votes. I am fine with others disagreeing with me, too. – Jack D'Aurizio Apr 18 '17 at 21:35

The series converges absolutely. This is because we have $$0 \leq \exp\left(-\frac{n^3}{3}\right) \leq \frac{1}{1 + \frac{n^3}{3} + \frac{n^6}{18}}$$

by considering the Taylor expansion and hence

$$0 \leq n^2\exp\left(-\frac{n^3}{3}\right) \leq \frac{n^2}{1 + \frac{n^3}{3} + \frac{n^6}{18}} \leq 18n^{-4}$$

and $$\sum 18n^{-4}$$ converges by the integral test.

More intuitively, just keep in mind that exponentials of the form $a^{-x}$ for $a>1$ decay faster than any rational function.

If you want to use the alternating series test to merely establish conditional convergence, you can use L'hopital to show $\lim_{x \to \infty} x^2e^{-\frac{x^3}{3}} = 0$

• Thank you I see how to solve the question much more clearly now. – goldenlinx Apr 18 '17 at 21:21

Letting $a_n = (-1)^n n^2 e^{-\frac{n^3}{3}}$, we have $$\frac{a_{n+1}}{a_n} = - \frac{(n+1)^2}{n^2} e^{-\frac{(n+1)^3}{3}}e^{\frac{n^3}{3}} = -\left(1+\frac{1}{n}\right)^2 e^{-\frac{3n^2+3n+1}{3}}$$

Then as $n \to \infty$, note $\left(1+\frac{1}{n}\right)^2 \to 1$ and since $n^2+n+\frac{1}{3} \to \infty$, the exponential tends to zero so $$\lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1$$ so the series converges by the ratio test.

$|a_n|=|(-1)^nn^2e^{-\frac {n^3}{3}}|=\left(n^2e^{-\frac {n^3}{6}}\right)e^{-\frac {n^3}{6}}\le e^{-\frac {n^3}{6}}\le e^{-n}\quad$ for $n$ large enough.

• The first inequality is because any polynom is small in regard to any exponential (their product is going to zero), so for $n\gg 1$ large enough, $n^2e^{-\frac {n^3}{6}}<1$.

• The second one is because $e^{-x}$ is decreasing, so since $\frac {n^3}6>n$, once again for $n\gg 1$ large enough, we have $e^{-\frac {n^3}{6}}\le e^{-n}$

So for $n$ large enough we have $|a_n|<e^{-n}$ which is a term of a convergent geometric serie (of reason $0<\frac 1e<1$), so the original serie is absolutely convergent (because positive terms makes the partial sum increasing and it is bounded by the sum of the other serie, so it has a limit).

Remember that if you can get a straightforward result by rough majorations like in this case, always go for it before invoking more powerful tools. Advanced tools will give you more precise estimations, but it's good to come back to the basics, analysis is the art of minoration and majoration.

The series is absolutely convergent. Apply the quotient criterion.