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

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 a similar looking series before? Thanks.

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In addition to the nice answers you received, you can do a quick check on WolframAlpha (Sum[n^(1/n) - 1,{n,1, Infinity}]) and it tells which tests it tried and which work to show divergence. Of course, you still need to determine why, but that might prove useful at times. – Amzoti Dec 7 '12 at 21:47

Rewrite $\sqrt[n]{n}$ as $n^{1/n}=e^{(\ln n)/n}$ and use $e^x-1\sim x$ to see that $$\sqrt[n]{n}-1\sim \frac{\ln n}{n}$$ (where I use $\sim$ to mean “is asymptotically equal to”). Now compare with the harmonic series.
A variation on Harald's answer. For all real $x>0$ and integral $n \geq 1$ $$\log(x) \leq n(x^{1/n} - 1)$$ and in particular $$\frac{\log(n)}{n} \leq n^{1/n} - 1.$$ So it is a divergent series.