Convergence of $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$

Question

Let $\lim _{n\to\infty}a_n=l$. Show that $\sum_{n=1}^\infty \frac{1}{n^{a_n}}$ converges if $l>1$ and diverges if $l<1$. 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 $p>1$ 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?

Answer 1

Hints: if $l > 1$ then proof that there is $N$ such that $a_n > 1$ for any $n>N$ same method for $l < 1$

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

Answer 2

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.

Answer 3

There are examples where $a_n \to 1$ and the series still converges. The trick is to make $a_n \to 1$ from above and make it approach $1$ very, very slowly. Indeed, take $$a_1 = 1,\,\,\,\,\, a_2 = 1, \,\,\,\,\, a_n = 1 + \frac{1}{\log(\log(n))}, \,\,\,\,\, \text{ for } \,\,\,\,\, n > e$$ and you'll find that $\sum^\infty_{n=1} \frac{1}{n^{a_n}}$ converges. You can prove this using the Cauchy Condensation test which says that for a positive decreasing sequence $b_n$, we have that $$\sum^\infty_{n=1} b_n \,\,\, \text{ converges iff } \,\,\, \sum^\infty_{n=1} 2^n b_{2^n} \,\,\, \text{ converges.}$$ Here take $b_n = \frac{1}{n^{a_n}}$ and note $$2^n b_{2^n} = 2^n \frac{1}{(2^{n})^{a_{2^n}}} = \frac{1}{(2^n)^{a_{2^n} - 1}}.$$ Ignoring the first couple terms, if $a_n = 1 + \frac{1}{\log(\log(n))}$, then we get $$2^n b_{2^n} = \frac{1}{2^{n/\log(\log(2^n))}} = \frac{1}{2^{n/\log(n\log(2))}}.$$ However, since any logarithm is asymptotically smaller than any power of $n$, we eventually find that $$\log(n\log(2)) \lesssim \sqrt n \,\,\, \implies \,\,\, \frac{1}{\sqrt n} \lesssim \frac{1}{\log(n\log(2))} \,\,\, \implies \,\,\, \sqrt n \lesssim \frac{n}{\log(n\log(2))}.$$ Thus, we have $$2^nb_{2^n}\lesssim \frac{1}{2^\sqrt{n}}.$$ For the latter, notice that for $n$ between $m^2$ and $(m+1)^2$, there are $2m+1$ terms and $2^{\sqrt n} \le 2^{\sqrt m}$ for all of them. Thus $$\sum^\infty_{n=1} \frac{1}{2^{\sqrt n}} \lesssim \sum^{\infty}_{m=1} \frac{2m+1}{2^m} < \infty$$ since the last series is just a differentiated geometric series. This shows that $\sum 2^nb_{2^n}$ converges by comparison and hence $\sum b_n$ converges by the condensation test.