finding values for absolute convergence

Question

Find all values of real number p or which the series converges:

$$\sum \limits_{k=2}^{\infty} \frac{1}{\sqrt{k} (k^{p} - 1)}$$

I tried using the root test and the ratio test, but I got stuck on both.

Answer 1

If $p\lt0$, then $\sqrt{k}(k^p-1)\sim-\sqrt{k}$ hence the series diverges. If $p=0$, then $k^p-1=0$ for every $k$ hence the series is undefined. If $p\gt0$, then $\sqrt{k}(k^p-1)\sim k^{p+1/2}$ hence the series converges if and only if $p\gt1/2$.

Answer 2

Let $p>1$, so since $$\lim_{k\to+\infty}k^p\frac{1}{\sqrt{k}(k^p-1)}<\infty$$ then for $p>1$ the series is convergent.

Answer 3

By cauchy's condensation test $$\frac{2^k}{2^{kp + k/2} - 2^{k/2}} = \frac{1}{2^{kp - k/2} - \frac{1}{2^{k/2}}} \simeq \frac{1}{2^{kp - k/2}}.$$ If we now use the ratio test for the series $\sum_k \frac{1}{2^{kp - k/2}}$ we obtain $$\frac{2^{kp - k/2}}{2^{kp - k/2} \cdot 2^{p - 1/2}} \to \frac{1}{2^{p-1/2}} < 1 \Longleftrightarrow p > 1/2.$$