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 $l>1$ 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?

• It's really only important to look at the tail of a series. You can use the comparison test on these tails and that will take care of the $\ell>1$ and $\ell<1$. The other answer below takes care of $\ell=1$ with a different comparison test. – Robert Wolfe Aug 21 '18 at 14:14
• According to WolframAlpha, you are right: if $a_n \to 1$ then it can either converge or diverge. Put $a_n = 1$ for all $n$ to see divergence (as you've said), or $a_n = 1 + \frac{1}{\log(\log(n))}$ to see convergence. I'm struggling to prove that this series converges though (and WA gives no real justification). – User8128 Aug 21 '18 at 14:34

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.

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.

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 $a_n = 1 + 1/n$

• "if $l > 1$ then proof that there is $N$ such that $a_n > 1$ for any $n>N$" Interestingly, the fact that $a_n > 1$ for every $n$ large enough, does not allow to conclude that the series converges. (@upvoter Care to explain your vote?) – Did Aug 21 '18 at 14:48
• in his question, its a fact that an converges to l where l > 1, so there is nothing wrong it what is written, for instance choose e = (l-1) / 2 – user3336593 Aug 21 '18 at 14:52
• @Did Thanks for the clarification, I was struggling with that. I would need that there is an $\epsilon>0$ and an $n\in \mathbb{N}$ such that $a_n\geq 1+\epsilon$ for all $n\geq N$ and then I can apply the comparison test right? Is this true that there is such an estimate? From my Intuition yes, because we can not have infintely many points which are arbitrarly closed to 1 but I can not make it precise... – mathstackuser Aug 21 '18 at 14:54
• @user3336593 You seem to be confusing the condition that $a_n>1$ for every $n$ large enough and the condition that there exists $c>1$ such that $a_n\geqslant c$ for every $n$ large enough. Your post mentions the former, which does not allow to conclude, while the latter does allow to conclude. – Did Aug 21 '18 at 14:55
• @mathstackuser Yes you need this condition. If $\ell>1$, it does hold, by the usual epsilon-N argument, for example with $c=\frac12(1+\ell)$, that is, $\epsilon=\frac12(\ell-1)$. – Did Aug 21 '18 at 14:56