investigate $\sum\limits_{n\ge1}{\frac{(-1)^n}{n^\alpha \ln n}}$

Question

I need to investigate the series (Hence, when the series converges and when the series converges absolutely depending on $\alpha$).

$$\sum\limits_{n\ge2}{\frac{(-1)^n}{n^\alpha \ln n}}$$

1. For $\alpha \gt 1$ the series converges because $a_n = \frac{1}{n^\alpha \ln n}$ is monotonic sequence converges to $0$ and therefore the sum $\sum (-1)^n a_n$ converges by Leibniz criterion (Alternating series test). Therefore, for $\alpha > 1$ the series converges and also converges absolutely.
2. The case where $\alpha \le 1$ is unclear to me. Basically, I understood I can use the fact that $\ln n < n^b$ for sufficiently large $n$ and for all $b$.

I'd be glad if you could guide me with #2 regarding both convergence and convergence absolutely.

Thanks.

Answer 1

The convergence of the series

• If $\alpha\ge0$ then by Leibniz theorem the series is convergent.
• If $\alpha<0$ then the series is divergent since the general term of this series doesn't even converge to $0$.

The absolute convergence of the series

• if $\alpha>1$ then the series is absolutely convergent by comparison with a convergent Riemann series: $$\frac{1}{n^\alpha \ln n}\le \frac{1}{n^\alpha }$$

• If $\alpha=1$ then by the integral test we see that the series isn't absolutely convergent.

• If $\alpha<1$ then the series isn't absolutely convergent by comparison with the last case.