# 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
• looks fine to me (+1) – tired Sep 20 '16 at 14:57
• for big enough $k$ you can just taylorexpand $\sqrt{k+1}-\sqrt{k}=\sqrt{1/k}+\mathcal{O}(1/k)$ to reach the same conclusion – tired Sep 20 '16 at 14:59
• Wow you see so fast, thank you! :D You know if there is an easier / faster way? – tenepolis Sep 20 '16 at 14:59
• have a look at my second comment – tired Sep 20 '16 at 15:00
• $<\sum \frac{1}{2k^2}=\frac{{\pi}^2}{3}$ – Takahiro Waki Sep 20 '16 at 15:08

$$\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}< \sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}$$ $$\sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}=\sum_{k=1}^{\infty}\frac{1}{k^{\frac{3}{2}}}=\zeta (\frac{3}{2})$$
• How would you continue after? I think there will be problems getting $k^{2}$ in the denominator? – tenepolis Sep 20 '16 at 15:01
• Doesn't matter since $\sum_k \frac{1}{k^p}$ converges when $p>1$. – Alexis Olson Sep 20 '16 at 15:02