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When I recently read about the power rule for infinite sums, I thought that the sum $$\sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$$ would converge, as it is akin to a $p$-series, but with an (admittedly nonconstant) $p=1+\frac{1}{n} \ge 1$, and so is strictly less than the harmonic series. However, when I ask WolframAlpha for an evaluation, it claims that the sum diverges by the comparison test. I could accept this, but for the fact that I am not aware of a particular sum strictly lesser than my above sum that diverges. What might be a sum that fulfills that criterion?

Lieutenant Zipp
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    Also: https://math.stackexchange.com/q/1752151/42969, https://math.stackexchange.com/q/2266680/42969, https://math.stackexchange.com/q/2054911/42969, https://math.stackexchange.com/q/1673312/42969. – Martin R Sep 01 '19 at 17:27
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    Oh. My question is not original. That’s disappointing. Oh well. What can you do? – Lieutenant Zipp Sep 01 '19 at 17:31
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    We have $n^{1/n}\le 2$, so $\frac 12\sum_{n=1}^N\frac 1n\le\sum_{n=1}^N\frac 1{nn^{1/n}}$. And there you have a series that is strictly less than your series. Both diverge. – amsmath Sep 01 '19 at 18:17

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