# Is $\sum_n\left(a_n^{\frac{1}{n}}-1\right)$ convergent?

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 $$\left( 1+\frac{1}{n}\right)^n for all $$n\geq 3$$. Hence $$\frac{1}{n} for all $$n\geq 3$$. So by comparison test the given series is divergent.

But I can not figure out the above question.

• For me your counterexample is right. Jul 23, 2020 at 6:25
• 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. Jul 23, 2020 at 7:40

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.