Converge or Diverge? Show that $\sum_{n=1}^{\infty} \frac{1}{n^{{n}/{\log(n)}}}$ converges 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
 A: @JBC and @robjohn dealt nicely with base $e$. 
In base $b$, $n^{n/\log_b(n)} = b^n$, since $n=b^{\log_b n}$. 
(That is, the exponential function is the inverse function of the logarithmic function.) 
In fact, $\lim_{n\to 1} n^{n/\log_b(n)} = b$, so the formula can be used for all relevant $n$. 
Thus, 
$$\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n^{n/\log_b(n)}} 
&=& \sum_{n=1}^\infty \left(\frac{1}{b}\right)^n.
\end{eqnarray*}$$
This is just a geometric series which converges to $\frac{1}{b-1}$ for $b>1$. 
A: 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.
A: Hint: $n^{\frac{n}{\log(n)}}=e^n$
Although the $n=1$ term will have to be handled with care :-)
A: METHOD I
It's easy to see that:
$$\sum_{n=1}^{\infty} \frac{1}{n^{\frac{n}{\log(n)}}}\leq\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}\leq 1+\sum_{n=2}^{\infty} \frac{1}{n(n-1)}=2$$ 
We may conclud that the sum converges.
METHOD II
Cauchy condensation test works pretty fast, as well. 
Q.E.D. (these are just 2 alternative ways to the geometric series)
