# Does absolute convergence of $\sum_{n=1}^{\infty} a_n$ implies $(a_n)^{1/n}$ tends to some $r \in [0.1]$?

True or false: Let $$\sum_{n=1}^{\infty} a_n$$ be an absolutely convergent series, then $$(a_n)^{1/n} \rightarrow r \in [0.1]$$?

My initial progress: evidently $$|a_n|\to 0$$,so eventually $$|a_n| ＜1$$ and therefore $$|a_n|^{1/n}＜1$$. I tried to use proof of contradiction at this point to show that if $$|a_n|^{1/n}$$ doesn't converge to anywhere in $$[0,1]$$ something will go wrong, but I didn't succeed...

This is essentially the converse of the root test. Therefore I went to several analysis book and look up for a potential counter-example, but I didn't find any relevant discussion.

If the statement in the question is true, supply a detailed proof. Otherwise, provide a counter-example.Thank you!

• @DavideGiraudo Hey, be kind, I'm new to the community and isn't very aware of the rules. I'll add on my thoughts and progress to the question now. Thanks. Jul 7 at 6:31
• @AnneBauval Thank you, I'm new to here and wasn't aware to the mistake. I apologize and I just added what I have on my scratch paper at the moment. Jul 7 at 6:37
• @RyanZhou No problem. I removed my downvote and converted it into an upvote. Jul 7 at 6:38
• Not going to work. If you have two series $a_n,b_n$ such that $\sum a_n < \infty, \sum b_n < \infty$ but $\lim_{n \to \infty} a_n^{\frac 1n}$ and $\lim_{n \to \infty} b_n^{\frac 1n}$ have different values (something like $a_n = 2^{-n}$, $b_n = 3^{-n}$) then the sequence $a_1,b_1,a_2,b_2,a_3,b_3,\ldots$ is summable without the required limit existing because it's different along the odd and even subsequences. If you insist that the sequence $a_n^{\frac 1n}$ converges, then the limit must lie in $[0,1]$. Jul 7 at 6:46
• Consider interleaving the terms of an absolutely convergent series for which the limit exists and is nonzero with an infinite sequence of zeros Jul 7 at 6:48

If $$\sum a_n$$ is absolutely convergent, the best you can say is (by the Cauchy–Hadamard theorem) $$\limsup_{n \to \infty} \left( |a_n|^{1/n} \right)=\frac1R\le1$$ but $$\left( |a_n|^{1/n} \right)$$ is not necessarily convergent.
An easy counterexample is $$a_{2n}=\frac1{n^2},\quad a_{2n+1}=0.$$
The thing here is that you can have a subsequence for which the root test works and an other one for which the root test fails (the limit of $$(a_n)^{1/n}$$ along this subsequence is $$1$$) but the series is still convergent.
This leads to the counter-example $$a_{2n}=2^{-2n}$$ and $$a_{2n+1}=n^{-2}$$.
• Your example works just fine. It's just that the question isn't quite about if the root test applies. But yes, the series formed by combining the sequences is absolutely convergent, but $|a_n|^{1/n}$ tends to 1/4 and 1 for the two subsequences. Many thanks. Jul 7 at 6:50