# Having trouble verifying absolute/conditional convergence for these series

Greetings,

I'm having trouble applying the tests for convergence on these series; I can never seem to wrap my head around how to determine if they're absolutely convergent, conditionally convergent or divergent.

a) $\displaystyle \sum_{k=1}^{\infty}\frac{\sqrt{k}}{e^{k^3}}$.

b) $\displaystyle \sum_{k=2}^{\infty}\frac{(-1)^k}{k(\ln k)(\ln\ln k)}$.

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## 3 Answers

Hint: I imagine that the Ratio Test is part of your toolbox. Problem (a) should yield fairly easily to that.

For (b), note that the given series is (almost) an alternating series. (The first two terms have the same sign.) To prove that the series does not converge absolutely, a tool that works is the Integral Test. Try differentiating $\ln(\ln(\ln x))$.

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2
I should have put in some heuristic comments. The Ratio Test is good if, informally, the terms decrease "real fast." As for (b), presumably it screams alternating series. And probably, in course notes or the book, $\sum 1/(k\ln k)$ was discussed. Imitate. – André Nicolas Apr 8 '11 at 20:06
Thanks for the hints, the community here is very helpful. I'm making progress, albeit slow progress. – sephy7324 Apr 8 '11 at 20:22

The first series is positive, so it either converges absolute, or diverges. My gut reaction was to try some back-of-the-envelope comparison.

Let's see: $\sum \frac{1}{e^{k}}$ converges, and $\frac{1}{e^{k^3}}\leq \frac{1}{e^k}$, so $\sum\frac{1}{e^{k^3}}$ converges. The exponential dominates $\sqrt{k}$, so personally, I would expect the series to converge. I would try an integral test, but it looks like an annoying function to integrate. So perhaps we can find a series $\sum\frac{p(k)}{e^{k^3}}$ for some function $p(k)$, such that $\int_1^{\infty}p(x)e^{-x^3}\,dx$ is easy to integrate.

This is doable, but as user6312 pointed out, the Ratio Test will do the job as well.

The second series is (eventually) alternating. Try the Alternating Series Test. To check the absolute convergence, try an Integral test. You'll need a change of variable. Hint. If you set $u=\ln x$, what happens?

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a) Since $e^x >x$ then $$\frac{\sqrt{k}}{e^{k^3}}<\frac{\sqrt{k}}{k^3}=\frac{1}{k^{5/2}}$$ But $\sum_{k=1}^\infty \frac{1}{k^{5/2}}$ is convergent ($p$-series test). Hence the original series is convergent (hence absolutely convergent since it is a positive series).

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