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.
Best Answer
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.