Determine for which the value of $\alpha $ of the given series is convergent

Question

Determine for which the value of $\alpha $ of the given series is convergent

$\sum_{n=1}^{\infty}(\sqrt [n] a-1)^\alpha $ where $a >1$.

I thinks if $\alpha $ $< 0$ then the given series $\sum_{n=1}^{\infty}(\sqrt[n] a-1)^\alpha $ will converge

becuse i know that $lim_{n\rightarrow \infty} \frac{\sqrt[n] a -1}{1/n} = loga $ where $a >1$

Answer 1

Note that

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

then

$$(\sqrt [n] a-1)^\alpha\sim \frac{\log^{\alpha} a}{n^{\alpha}}$$

and the given series converges for $\alpha>1$ by limit comparison test with $\sum \frac 1 {n^p}$ for $1<p<\alpha$ and diverges otherwise.

Answer 2

The series converges if and only if $\alpha >1$. Compare the given series with $\sum_{n=1}^{\infty} (\frac 1 n \ln (a))^{\alpha}$ which converges iff $\alpha >1$.