I know that $$\sum_{n=1}^{\infty}{n^{1/n}}$$ diverges, since the limit when $n$ tends to infinity of $n^\frac{1}{n}$ is equal to $1$. With this, I cannot conclude anything about the convergence of $$\sum_{n=1}^{\infty}\frac{1}{\sum_{k=1}^{n}k^{1/k}}$$ I don't see how to apply the root test, and I don't know which series I should choose to apply the comparison test. If I apply the Cauchy Condensation test we have $$\sum_{n=1}^{\infty}\frac{2^n}{\sum_{k=1}^{2^n}k^{1/k}}$$ and i don't know how to solve that limit
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1do you mean $\sum_{n=1}^{\infty}\frac{1}{\sum_{k=1}^{n}k^{1/k}}$? – timon92 Feb 06 '24 at 17:07
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I want to write this: $$\sum_{n=1}^{\infty}\frac{1}{n+n^1/2+...+n^1/n}$$. – Rúben Reis Feb 06 '24 at 17:11
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it is $\sum_{n=1}^{\infty}\frac{1}{\sum_{k=1}^{n}n^{1/k}}$ then. – timon92 Feb 06 '24 at 17:20
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Another one: https://math.stackexchange.com/q/3429545/42969 – Martin R Feb 06 '24 at 17:29