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Let $\{a_n\}$ is a sequence of positive real numbers and $\lim\limits_{n\rightarrow \infty}a_n^{\frac{1}{n}}=1$ , then is there any condition on $\{a_n\}$ so that the $\sum_n\left(a_n^{\frac{1}{n}}-1\right)$ is convergent $?$

I have seen one related question that : Prove or disprove that $\sum_n \left( n^{\frac{1}{n}}-1\right)$ is convergent.

My Approach is : $\lim\limits_{n \rightarrow \infty} \left( 1+\frac{1}{n}\right)^n=e$, so $\left( 1+\frac{1}{n}\right)^n<n$ for all $n\geq 3$. Hence $\frac{1}{n} <n^{\frac{1}{n}}-1$ for all $n\geq 3$. So by comparison test the given series is divergent.

But I can not figure out the above question.

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  • $\begingroup$ For me your counterexample is right. $\endgroup$
    – Michael Rozenberg
    Jul 23, 2020 at 6:25
  • $\begingroup$ With $b_n = a_n^{1/n}$ your question is equivalent to asking which conditions on a sequence $(b_n)$ with $b_n \to 0$ guarantee that $\sum b_n$ is convergent. $\endgroup$
    – Martin R
    Jul 23, 2020 at 7:40

1 Answer 1

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I suppose that $\ \forall n \in \mathbb N \ , \ a_n\geqslant 1$.
Then: $\ \sum_n \left(a_n^{\frac{1}{n}} -1\right) \ $ is convergent if and only if $\ \sum_n\dfrac{\ln(a_n)}{n} \ $ is convergent.

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