Do $\sum _{n=1}^{\infty }\:\left(\frac{1}{n\:}\right)^{\ln\:n}$ and diverge?

Question

So I want to find out if $\sum _{n=1}^{\infty }\:\left(\frac{1}{n}\right)^{\ln n}$ and $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\ln\left(2+\ln n\right)}$ diverges or not.

For $\sum _{n=1}^{\infty }\:\left(\frac{1}{n }\right)^{\ln n}$, I see that I have $\frac{1}{n}$ and that it might help, but I have no clue of how to get rid of $\ln(n)$. So I am a bit stuck here.

For $\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\ln\left(2+\ln n\right)}$, I would like to think that the series are alternating. Also, since the coefficient $\ln (2+\ln n)$ is monotonically increasing, the series is not convergent according to Leibniz (I think?).

Can I say that because $\sum _{n=1}^{\infty }\:\frac{1}{n}$ is divergent, my series is also divergent then?

Answer 1

1. If $\ln n > 2$, then $\frac{1}{n^{\ln n}}<\frac{1}{n^2}$. What can you deduce from this?

2. The sequence $\frac{1}{\ln(2+\ln n)}$ is monotonically decreasing to zero as $n\to\infty$. Is there a theorem you know that can help you decide what happens when a monotonically decreasing to zero sequence appears with changing signs in a series?

Answer 2

For 2), use alternating series test. The summand converges to $0$. To show monotonicity, prove that the summand grows approximately as $\frac{1}{\log \log n}$.

For 1), note that the summand can be written as $e^{-\log^2n} $, so it can be compared to the integral $\int_{0}^{\infty}e^{-\log^2 x}dx$. Setting $\log x =t$, we get $\int_{0}^{\infty}e^{-t^2 + t}dt = e^{\frac{1}{4}} \int_{0}^{\infty}e^{-(t-\frac{1}{2})^2}dt = \frac{e^{\frac{1}{4}}\sqrt{\pi}}{2}$. Hence your series converges