Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does the following series converge? $$\sum_{n=1}^\infty \frac{1}{n\sqrt[n]{n}}$$

As $$\frac{1}{n\sqrt[n]{n}}=\frac{1}{n^{1+\frac{1}{n}}},$$ I was thinking that you may consider this as a p-series with $p>1$. But I'm not sure if this is correct, as with p-series, p is a fixed number, right ? On the other hand, $1+\frac{1}{n}>1$ for all $n$. Any hints ?

share|cite|improve this question
you could use root test or ratio test to get a hint. –  mez Feb 11 '13 at 14:24
did you miss n square root in the base? –  mez Feb 11 '13 at 14:25
If the series is $\sum 1/n\sqrt{n}$ as written, it is a $p$ series with $p=3/2$ so converges. But after "As...I was thinking" it looks like the series is different. Was it supposed to be $\sum 1/(n \cdot n^(1/n))$ for the problem? –  coffeemath Feb 11 '13 at 14:26
sorry to be confusing, fixed it, it's about this series $\sum_{n=1}^\infty \frac{1}{n\sqrt[n]{n}}$ –  Kasper Feb 11 '13 at 14:27

4 Answers 4

up vote 16 down vote accepted

Hint: $\sqrt[n]{n}\to1$ when $n\to\infty$ hence, by comparison with the series $\sum\limits_n\frac1n$, this series $______$.

share|cite|improve this answer
How do you prove $\sqrt[n]n \to 1$ ? –  Kasper Feb 11 '13 at 14:29
Sub-hint: logarithm. –  Did Feb 11 '13 at 14:30
To prove $n^{\frac{1}{n}}\rightarrow 1$, first notice that $n^{\frac{1}{n}}=1+\epsilon$ and then use binomial theorem. –  user60469 Feb 11 '13 at 14:39
@Did hm... I get $\sqrt[n] n \leq e^{\frac{1}{\sqrt n}}$ for sufficiently large $n$ –  Kasper Feb 11 '13 at 14:42
No idea how you got that. Try using $\sqrt[n]{n}=\exp((\log n)/n)$ and the fact that $(\log n)/n$ has a known limit when $n\to\infty$. –  Did Feb 11 '13 at 14:44

Note that $\sqrt[n]{n}\le 2$. This can be proved by induction, for it is equivalent to $n\le 2^n$.

Thus $$\frac{1}{n\sqrt[n]{n}}\ge \frac{1}{2n}.$$ It follows by Comparison with (half of) the harmonic series that our series diverges.

share|cite|improve this answer
Aah, this is a very neat way as well ! –  Kasper Feb 13 '13 at 15:41
+1 very simple ! –  pbs Jul 23 at 15:30

Limit comparison test:


So that both

$$\sum_{n=1}^\infty\frac{1}{n\sqrt[n] n}\,\,\,\text{and}\,\,\,\sum_{n=1}^\infty\frac{1}{n}$$

converge or both diverge...

share|cite|improve this answer

AM-GM gives $$ \sqrt[n]{n}\leq\frac{1}{n}(n+\underbrace{1+\cdots+1}_{n-1})=\frac{1}{n}(2n-1)<\frac{1}{n}2n\implies\frac{1}{n\sqrt[n]{n}}>\frac{1}{2}\frac{1}{n} $$ so $$ \sum_{n=1}^M\frac{1}{n\sqrt[n]{n}}>\frac{1}{2}\left(\sum_{n=1}^M\frac{1}{n}\right)\cdot $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.