# Does $\sum_{k=1}^{\infty} \frac 1 {k^{1+\frac 1 k}}$ converge or diverge? [duplicate]

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?

## 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.

• – Martin R Aug 3 '18 at 20:24
• For $k\ge1$ $$\frac{1}{k^{1+1/k}}\ge \frac{1-\frac1k\log(k)}{k}=\frac1k - \frac{\log(k)}{k^2}$$ – Mark Viola Aug 3 '18 at 20:34

## 2 Answers

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.

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