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This question already has an answer here:

$\sum_{k=1}^{\infty} \frac 1 {k^{1+\frac 1 k}} $

What I've tried:

$a = $ $\sum_{k=1}^{\infty} \frac 1 {k^{1+\frac 1 k}} $

$b = $ $\frac 1 k$

Limit comparison test:

$\lim_{n\to {\infty}} \frac a b = 1$

Therefore, by Limit comparison test, a and b diverge or converge together.

Because b diverges (p-series), a must also diverge.

Have I made any mistakes?

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marked as duplicate by Martin R, Community Aug 3 '18 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Yes you are absolutely right indeed

$$\frac{\frac 1 {k^{1+\frac 1 k}}}{\frac1k}=\frac k {k^{1+\frac 1 k}}=\frac 1 {k^{\frac 1 k}}\to 1$$

and by limit comparison test we can conclude.

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Since $k^{1/k}\sim 1$, $k^{-1-1/k}\sim k^{-1}$ so the series diverges.

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