Does $\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k\ln k}$ converge?

Question

let $a_k=\dfrac{1}{k\ln k}$

I found out that $a_{n+1} < a_n$ for all $n$.

Also, $\displaystyle\lim_{n\to \infty} a_n = 0$.

So by leibniz alternating series test, I said $\displaystyle\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k\ln k}$ converges.

Is that true? Does it Absolute converge?

Answer 1

As you noted, it converges by Leibniz test. However, it does not converge absolutely. (Try the integral test on $\sum |a_k|$.)

Answer 2

It does converge because it is alternating and elements decrease monotonically to 0. It does not converges absoluteley, see the condensation test.

I hope this helps ;-)