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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

For any $k\in \{1,2,...,n\}$ it holds that $n^{1-\frac{1}{n}}\ge n^{1-\frac{k}{n}}$. Because of that: $$n^{2-\frac{1}{n}}=n\cdot n^{1-\frac{1}{n}} \ge n^{1-\frac{1}{n}}+ n^{1-\frac{2}{n}}+\dots+ n^{\frac{1}{n}} +1$$ Then: $$\sqrt[n]{n}-1=\frac{n-1}{n^{1-\frac{1}{n}}+ n^{1-\frac{2}{n}}+\dots+ n^{\frac{1}{n}} +1}\ge \frac{n-1}{n^{2-\frac{1}{n}}} \sim \frac{n}{n^{2-\frac{1}{n}}}=\frac{\sqrt[n]{n}}{n}\ge \frac{1}{n}$$ Then using comparison tests (both direct and limit comparison test) the series diverges.