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Let be $u_n = \dfrac{1}{\sum\limits_{k=1}^n k^{\frac{1}{k}}}$, I am trying to show that $\sum\limits_{n \geq 1} u_n$ is divergent.

First, I tried to naive ideas, comparing it to a usual Riemann series, applying Alembert / Cauchy rules and doing series expansion.

I am wondering if it has to do with Cesaro theorem.

StubbornAtom
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Raito
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    @AlexR.: the denominator converges?! $\sum_{k=1}^{n}k^{1/k}$ behaves more or less like $n$, the point is to understand how much such sum is bigger than $n$. Something like $n$ implies divergence, something like $n\sqrt{n}$ implies convergence, so yours is not a strong hint. – Jack D'Aurizio Oct 26 '17 at 18:14
  • @JackD'Aurizio: Whoops, misread the exponent. – Alex R. Oct 26 '17 at 18:23

3 Answers3

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We know that $k^{1/k}\to 1$ as $k\to\infty$. This also implies that $$ \frac{1}{n}\sum_{k=1}^n k^{1/k} \to 1$$ (it's a general fact, if the limit of a sequence exists, then also the cesaro average converges to it).

In particular this implies that $\sum_{k=1}^n k^{1/k}\sim n$ and so $u_n\sim\frac{1}{n}$. But then we can conclude that $\sum_{n\geq 1} u_n $ diverges since $\sum_{n\geq 1}\frac{1}{n}$ does.

Lucio
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By the AM-GM inequality, for any $k\geq 2$ we have $$ k^{1/k}=\text{GM}\left(1,\frac{2}{1},\ldots,\frac{k}{k-1}\right)\leq \text{AM}\left(1,\frac{2}{1},\ldots,\frac{k}{k-1}\right)=1+\frac{H_{k-1}}{k} $$ hence it follows that for any $n\geq 2$ we have $$ \sum_{k=1}^{n}k^{1/k} \leq n+\sum_{k=1}^{n-1}\frac{H_{k}}{k+1}\stackrel{\text{SBP}}{=}n+H_{n-1}(H_n-1)-\sum_{k=1}^{n-2}\frac{H_{k+1}-1}{k+1} $$ and $$ \sum_{k=1}^{n} k^{1/k} = n +O(\log^2 n) $$ trivially implies that the original series is divergent.

Jack D'Aurizio
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We have

$$k^{1/k} =e^{\frac{\log k}k}=1+\frac{\log k}k+o\left(\frac{\log k}k\right)$$

and using that result

$$\sum_{n=1}^{n}k^{1/k}\sim n+\frac{\log^2 n}2$$

therefore by limit test $\sum_{n=1}^{\infty}\frac{1}{\sum_{k=1}^{n}k^{1/k}}$ diverges.

user
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