A student was recently asked this question by his instructor:


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


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


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.

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Thank you! This is very helpful. $\endgroup$ – user52183 Dec 7 '12 at 21:39

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