Why does $\sum_{n=1}^\infty (\sqrt[n]{a} - 1)$ diverge for $1 \neq a>0$?

Question

Why does $\sum_{n=1}^\infty (\sqrt[n]{a} - 1)$ diverge for $1 \neq a>0$?

We tried to proof that $\sqrt[n]{a} - 1 > 1/n$, but this doesn't hold. Any ideas?

Answer 1

Assume $a>1$ and note that $$\bigl(a^{1/n}-1\bigr)\ \sum_{k=0}^{n-1}a^{k/n}=a-1\ .$$ From $a^{k/n}<a$ $\ (0\leq k<n)$ it follows that the sum is $<n\> a$ and therefore that $$a^{1/n}-1>{a-1\over a}\ {1\over n}>0\ .$$ Similarly, when $0<a<1$ then $a^{k/n}\leq 1$ $\ (0\leq k<n)$. Therefore $\sum_{k=0}^{n-1}a^{k/n}<n$, and it follows that $$1-a^{1/n}>(1-a)\ {1\over n}>0\ .$$ In both cases it follows from the divergence of the harmonic series that the considered series $\sum_{n=1}^\infty \bigl(a^{1/n} -1\bigr)$ diverges.

Answer 2

If $a>0$ and $a \ne 1$, then $\sqrt[n]{a}-1 = e^{\frac{\log{a}}{n}}-1 \sim \frac{\log a}{n}$ whose series clearly diverges.

Answer 3

In line with @Romeo’s answer, it happens that $$ \log(a)=\lim_{n\to\infty}n\left(\root{n}\of{a}-1\right) $$ so that $\root{n}\of{a}-1$ gets to look very much like $\frac1n\log a$.