<s>[INST] Prove or disprove: $\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}$ is convergent 
Prove or disprove: The following series is convergent
$$\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}$$

$$\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}= \frac{\left(\sqrt{k+1}-\sqrt{k}\right)\cdot \left(\sqrt{k+1} + \sqrt{k}\right)}{k\sqrt{k} \cdot \left(\sqrt{k+1} + \sqrt{k}\right)}= \frac{k+1-k}{k\sqrt{k} \cdot \left(\sqrt{k+1}+\sqrt{k}\right)}$$
$$=\frac{1}{k\sqrt{k} \cdot \left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{1}{\left(k\sqrt{k}\right)\cdot \left(\sqrt{k+1}\right)+k\sqrt{k}\cdot\sqrt{k}}= \frac{1}{k\sqrt{k}\cdot \left(\sqrt{k+1}\right)+k^{2}}< \frac{1}{k^{2}}$$
$$\Rightarrow\sum_{k=1}^{\infty}\frac{1}{k^{2}}$$
This is a convergent series and thus the original series is convergent as well.



*

*Did I do everything correcty (I'm especially not sure about the last step where I used "<")?

*Is there another way of proofing convergence here without that much work? I have tried ratio test too but it got so complicated and I couldn't solve it

 [/INST] $$\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}< \sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}$$
$$\sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}=\sum_{k=1}^{\infty}\frac{1}{k^{\frac{3}{2}}}=\zeta (\frac{3}{2})$$
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