Convergence of $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$ Let $\lim _{n\to\infty}a_n=l$. Show that $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$ converges if $l&gt;1$ and diverges if $l&lt;1$. What happens if $l=1$?

I tried to use the ratio test, but could not get a good estimate, I have difficulties, that there is the series $a_n$ involved. I know that $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if $p&gt;1$ and diverges if $p\geq 1$ but I am not sure how to use this exactly. 

For $l=1$, I guess both things could happen? Definitely, we can take $a_n=1$ for all $n$ and then we get the harmonic series which is divergent. Is there an example where the series converges?
If $l=1$ nothing can be said, since $\sum_{n\geq 1}\frac{1}{n\log^2(n+1)}$ is convergent but $\sum_{n\geq 1}\frac{1}{n\log(n+1)}$ is divergent by Cauchy's condensation test.
Hints:
if $l &gt; 1$ then proof that there is $N$ such that $a_n &gt; 1$ for any $n&gt;N$ 
same method for $l &lt; 1$

for $l = 1$
consider the series $a_n = 1 + 1/n$