10

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 ?

Did
  • 279,727
Kasper
  • 13,528
  • 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 Answers4

18

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

Did
  • 279,727
  • 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
  • 4
    To prove $n^{\frac{1}{n}}\rightarrow 1$, first notice that $n^{\frac{1}{n}}=1+\epsilon$ and then use binomial theorem. –  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
  • 2
    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
  • @Did I got my result as $(\log n)/n \leq 1/ \sqrt n$. Using the squeeze theorem I get the limit is indeed 1. This means the series must diverge right ? Thanks for all the help ! – Kasper Feb 11 '13 at 15:45
11

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.

André Nicolas
  • 507,029
6

Limit comparison test:

$$\frac{\frac{1}{n\sqrt[n]n}}{\frac{1}{n}}=\frac{1}{\sqrt[n]n}\xrightarrow[n\to\infty]{}1$$

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...

Joe
  • 4,757
  • 5
  • 35
  • 55
DonAntonio
  • 211,718
  • 17
  • 136
  • 287
2

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 $$ \sum_{n=1}^M\frac{1}{n\sqrt[n]{n}}>\frac{1}{2}\left(\sum_{n=1}^M\frac{1}{n}\right)\cdot $$

Kim Jong Un
  • 14,794
  • 1
  • 24
  • 49