# Series Converge or Diverge

• Dec 6th 2012, 09:37 PM
fatpolomanjr
Series Converge or Diverge
A student was recently asked this question by his instructor:

$\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}$

Converge or diverge?

I feel a little dumb for not being able to answer it. The following tests fail to prove convergence or divergence:

nth term test for divergence (limit is 0)
ratio test (limit is 1)
root test (see ratio test)
limit comparison with $\sqrt[n]{n}$ (not sure why I thought that'd work)

Something I did try was using the fact that

$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})$,

to rewrite $\sqrt[n]{n}-1$ as

$\frac{n-1}{n^{1-1/n}+n^{1-2/n}+\cdots+n^{2/n}+n^{1/n}+1}$.

However, I'm not sure what to compare this to. According to wolfram alpha this series "diverges by the comparison test", but comparison to what? There is a similar problem in Baby Rudin, but for $(\sqrt[n]{n}-1)^n$, and a simple nth root test resolves that series [convergence] in a hurry. Any ideas? Have any of you encountered such an easy looking series before? Thanks.
• Dec 7th 2012, 05:31 PM
Zouzz
Re: Series Converge or Diverge
There you are, I hope it helps :D
• Dec 7th 2012, 11:18 PM
fatpolomanjr
Re: Series Converge or Diverge
Thank you for this!

I found the answer through another recently, though. One way to show divergence was to use the inequality $\ln(x)\leq n(x^\frac{1}{n}-1)$ for all $x\geq1$ and $n\geq1$. Then substituting $n$ gives

$\frac{1}{n}\leq\frac{\ln(n)}{n}\leq \sqrt[n]{n}-1$.

Therefore the series diverges. Your solution is very nice because it gives a direct verification of the necessary inequality for the comparison test, which I find a bit more intuitive than "find the right Analysis inequality to get the job done" kind of argument.