Is convergent or divergent $\sum_{n=1}^\infty{(-1)^n\dfrac{\ln{n}}{n\ln{\ln{n}}}}$?

Question

$$\sum_{n=1}^\infty{(-1)^n\dfrac{\ln{n}}{n\ln{\ln{n}}}}$$

Any suggestions? I tried absolute convergence, but it doesn't work.

Answer 1

For $n=1$ $\log\log n=-\infty$, or rather it is not defined as it would have to be $\log\log1=\log0$. So the sum should start from 2.

I would try absolute convergence, only to find this does not converge absolutely as it is at least as big as the harmonic series ($\log\log n\leq\log n$ so that fraction is at least $\frac{\log n}{n\log n}=\frac1n$).

Then Leibniz, perhaps? As André notes, the absolute values are eventually decreasing (or non-increasing, though here the monotonicity is strict) since they are the product of $\frac{\log n}{n}$ and $\frac{1}{\log\log n}$, which are eventually decreasing, the second one since $\log\log n$ is evidently increasing, and the first one because it has derivative $\frac{1}{n^2}+\frac{\log n}{-n^2}$ which is negative. Now Leibniz demands three conditions:

1. That the terms $a_n$ for which the series $\sum(-1)^na_n$ is being examined be nonnegative, and $\frac{\log n}{n\log\log n}$ is evidently nonnegative, in fact, for $n\geq2$, strictly positive, and for $n=1$ it is undefined;
2. That those $a_n$ tend to 0, which is also evident since $\frac{1}{\log\log n}$ and $\frac{\log n}{n}$ both go to 0;
3. That those $a_n$ be eventually nonincreasing, as we proved above.

Hence, by Leibinz, that series converges, provided the sum starts from 2 and not 1, since at 1 it has an undefined term.

Answer 2

It is convergent by Dirichlet's test, since $\{(-1)^n\}_{n\geq 2}$ is a sequence with bounded partial sums and $\left\{\frac{\log n}{n \log\log n}\right\}_{n\geq 2}$ is a sequence definitely decreasing to zero.