<s>[INST] Determine whether $\sum_{k=1}^{\infty}\frac{k+2}{\sqrt{k^5+4}}$ converges or diverges Question

Determine if
$$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}}$$
converges or diverges.

I'm looking for a proof that this sum converges in a simpler way than I've shown.
My (ugly) work
I have the following chain of implications for $k\ge 1$:

*

*$k^3(3k^2 - 4k -4) \ge -16 \Longrightarrow$

*$3k^5 + 16 \ge  4k^4 + 4k^3 \Longrightarrow$

*$4k^5 + 16 \ge k^5 + 4k^4 + 4k^3 \Longrightarrow$

*$2(k^5+4)^{1/2} \ge k^{5/2} + 2k^{3/2} \Longrightarrow$

*$2k(k^5+4)^{1/2} \ge (k+2)k^{5/2} \Longrightarrow$

*$\frac{2k}{k^{5/2}} \ge \frac{k+2}{\sqrt{k^5+4}}\Longrightarrow$

*$2\sum_{k=1}^{\infty} k^{-3/2} \ge \sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}}$.

Since the sum on the left-hand side converges, the given sum also converges.
Gross!
 [/INST] $$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}} \le \sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5}} \le 3+\sum_{k=2}^\infty\frac{2k}{k^{2.5}}=3+2\sum_{k=2}^\infty\frac1{k^{1.5}}$$
Hence it converges.
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