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

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?

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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.

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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.

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Thank you! But is there another possibility to proof that without the log? We didn't introduce the log formally yet. – mjb Jan 4 '13 at 14:26
You are welcome. Anyway, I don't think there is another simple approach which does not make use of $\log$. But this is just my opinion. – Romeo Jan 4 '13 at 14:39
Ok, but what is the precise argumentation here? We have $\sqrt[n]{a} -1 > \frac{\log{a}}{n} -1$, which is not $> C \frac{1}{n}$ or anything, right? – mjb Jan 4 '13 at 15:31

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 $\frac1n\log a$.

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Or to be more precise: $$\root{n}\of{a}-1 \geq \frac{\log(a)}{n}$$ for all $n > 0$. – WimC Jan 4 '13 at 14:23
@WimC: How did you get rid of the $-1$ on the right side? – mjb Jan 4 '13 at 15:15