For what values of p does this series converge

Question

$$\sum_{k=1}^∞\frac{k}{\sqrt{4+k^p}}$$

I've already tried the integral test but failed to find a primitive, and also check if $$\lim_{n\to∞}an=0$$ But not totally satisfied with the result

Answer 1

The series $\sum\limits^{\infty}_{k=1}\frac{1}{k^{s}}$ converges if and only if $s>1$.

In your case note that $4+k^{p} > k^{p}$, so since $\sqrt{}:[0,\infty) \to \mathbb{R}$ is monotonic also $$ \sum\limits^{\infty}_{k=1}\frac{k}{\sqrt{4+k^p}} \leq \sum\limits^{\infty}_{k=1}\frac{k}{\sqrt{k^{p}}} = \sum\limits^{\infty}_{k=1}k^{1-\frac{p}{2}}. $$ Therefore the series converges iff $\frac{p}{2}-1>1$, so if and only if $p>4$.

Answer 2

Let $p >0$. $\frac 1 {\sqrt 2 k^{p/2 -1}} \leq \frac k {\sqrt {4+ k^p}} \leq \frac 1 { k^{p/2 -1}}$ for $k$ sufficiently large. Comparing with $\sum \frac 1 { k^{p/2 -1}}$ we see that the series converges iff $p >4$. I leave the case $p \leq 0$ to you.