Is the series $\sum_{n=1}^\infty (\sqrt[n]{a}-1)$ convergent or divergent?

Question

Given $a>1$, establish whether the series $$\sum_{n=1}^\infty \left(\sqrt[n]{a}-1\right)$$ converges or diverges.

I think that it diverges. I want to expand $\sqrt[n]{a}-1$ using Taylor around $0$ but I don't see how to do it.

Answer 1

Note that

$$\sqrt[n]{a}-1=e^{\frac{\log a}{n}}-1\sim\frac{\log a}{n}$$

thus the series diverges by limit comparison test with $\sum\frac{1}{n}$.

or as an alternative by Bernoulli

$$\sqrt[n]{a}-1=\sqrt[n]{1+c}-1>1+ \frac c n-1=\frac c n$$

and then the series diverges also by comparison with $\sum\frac{1}{n}$.

Answer 2

$$\lim\limits_{n\to\infty}n(\sqrt[n]{a}-1)=\ln a$$ thus the series diverges by comparison with $\sum\frac{1}{n}$.

Answer 3

If $c>d>0$ we have $$ c^n-d^n \leq n(c-d)c^{n-1},\qquad c-d\geq\frac{c^n-d^n}{n c^{n-1}} $$ hence by assuming $a>1$ and by letting $c=\sqrt[n]{a}, d=1$ we get $$ \sqrt[n]{a}-1 \geq \frac{a-1}{n a^{\frac{n-1}{n}}}\geq\left(1-\frac{1}{a}\right)\frac{1}{n} $$ hence the given series is divergent by comparison with the harmonic series.