[Math] Converge or Diverge? Show that $\sum_{n=1}^{\infty} \frac{1}{n^{{n}/{\log(n)}}}$ converges

Question

The series

$$\sum_{n=1}^{\infty} \frac{1}{n^{{n}/{\log(n)}}}$$

converges according to Wolframalpha.

Now I am not sure what the best technique is handling this one. I am thinking about a comparison test.

Here is what I thought $n \geq 1 \iff \log(n) \geq 1 \iff \dfrac{n}{\log(n)} \geq 1 \iff n^{\dfrac{n}{\log(n)}} \geq 1 \iff 0 \leq \frac{1}{n^{\frac{n}{\log(n)}}} \leq 1 \iff \sum_{n=1}^{\infty} 0 \leq \sum_{n=1}^{\infty} \frac{1}{n^{\frac{n}{\log(n)}}} \leq \sum_{n=1}^{\infty} 1 $

So by the Comparison Test, it converges. Or I guess i "sqqqqququuuuzed" the sum =)

Now my concern is that my sum is bounded, but I guess that doesn't imply the sum exist because something like $\sin(n)$ diverges even though it is bounded. Any insights?

EDIT: $\log(n)$ isn't the natural log

Answer 1

When $n>1$, $n^{\frac{n}{\log n}}=e^{\frac{n}{\log n}\log n}=e^n$.

Then it's a geometric series : $$\sum_{n=2}^\infty\left(\frac{1}{e}\right)^n$$ with $\frac{1}{e}<1$.

So the series conveges.