$\limsup_{n\rightarrow\infty} a_n^{1/\log n}<1/e$ and $a_n>0$ then $\sum a_n$ converges If $\limsup_{n\rightarrow\infty} a_n^{1/\log n}<1/e$ and $a_n>0$ then $\sum_{n=1}^{\infty} a_n$ converges.
$$0<a_n < e^{-\log n}=\frac{1}{n}$$
Also, $\exp\{\frac{1}{\log n}\log a_n \} \le \exp(-1)$ so $\displaystyle\frac{\log a_n}{\log n}\le -1$.
How does the limsup help here?

You are close. Let the $\limsup$ be $\frac{1}{e^{1+\delta}}$. Then for large enough $n$ we have 
$$\log a_n\lt -(1+\delta/2)\log n.$$ 
That means that $a_n \lt \frac{1}{n^{1+\delta/2}}$. Now Comparison does it. 
