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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
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    $\begingroup$ looks fine to me (+1) $\endgroup$
    – tired
    Sep 20, 2016 at 14:57
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    $\begingroup$ 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 $\endgroup$
    – tired
    Sep 20, 2016 at 14:59
  • $\begingroup$ Wow you see so fast, thank you! :D You know if there is an easier / faster way? $\endgroup$
    – tenepolis
    Sep 20, 2016 at 14:59
  • $\begingroup$ have a look at my second comment $\endgroup$
    – tired
    Sep 20, 2016 at 15:00
  • $\begingroup$ $<\sum \frac{1}{2k^2}=\frac{{\pi}^2}{3}$ $\endgroup$ Sep 20, 2016 at 15:08

1 Answer 1

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

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  • $\begingroup$ How would you continue after? I think there will be problems getting $k^{2}$ in the denominator? $\endgroup$
    – tenepolis
    Sep 20, 2016 at 15:01
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    $\begingroup$ Doesn't matter since $\sum_k \frac{1}{k^p}$ converges when $p>1$. $\endgroup$ Sep 20, 2016 at 15:02
  • $\begingroup$ Ohhh that's very important info for me, didn't know that, thanks so much! $\endgroup$
    – tenepolis
    Sep 20, 2016 at 15:03
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    $\begingroup$ @tenepolis according to Reimann criteria the series is converge $\endgroup$
    – E.H.E
    Sep 20, 2016 at 15:05
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    $\begingroup$ We didn't have Reimann in our readings, do you think we are allowed to use it in the exam then? Because if I know how to use it, why should I be forbidden to do it? You know or you think I shouldn't better do it? It saves a lot time for sure.. $\endgroup$
    – tenepolis
    Sep 20, 2016 at 15:09

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