I have to prove that the following series converges but not absolutely:
$$\sum_{k=0}^\infty (-1)^k\ \left(\sqrt{k+1}-\sqrt{k}\right)$$

I have used the Leibniz test (alternating series test) to prove that the series converges. Now how do I prove it does not converge absolutely?

  • $\begingroup$ What does k run from? $1$ to $\infty$? $\endgroup$
    – George
    Feb 12 '16 at 21:12
  • $\begingroup$ the summation runs from 0 to ∞ @George $\endgroup$
    – Room
    Feb 12 '16 at 21:16
  • $\begingroup$ The series diverges. Perhaps you should check the source to see if you have copied it correctly. $\endgroup$ Feb 12 '16 at 21:32
  • 1
    $\begingroup$ Another simple way of showing divergence of the absolute-value series without telescoping: $(\sqrt{k+1}-\sqrt{k})(\sqrt{k+1}+\sqrt{k}) = (k+1)-k = 1$, so $\sqrt{k+1}-\sqrt{k} = \frac{1}{\sqrt{k+1}+\sqrt{k}}$ $\geq\frac{1}{2\sqrt{k+1}}$. But now the sum of the latter diverges by comparison with e.g. the harmonic series. $\endgroup$ Feb 12 '16 at 21:48

Hint: Recall telescoping series

Answer: $$\sum_{k=0}^N | (-1)^k \left(\sqrt{k+1}-\sqrt k\right) | = - \sum_{k=0}^N \left(\sqrt{k} - \sqrt{k+1} \right) = \\ -\left(\sqrt{0}- \sqrt 1 + \sqrt{1} - \sqrt{2} + \sqrt{2} - \sqrt{3} + \dots - \sqrt{N} + \sqrt{N} - \sqrt{N+1}\right) = \sqrt{N+1}, $$

which does not converge as $N\to \infty$.

  • $\begingroup$ not really. Should I just try to write down the first few terms and see where that goes or what do u mean? I'm also not sure what telescoping series means. I am learning this in another language than English so the series may have a different name. Can you please be more specific ? $\endgroup$
    – Room
    Feb 12 '16 at 21:25
  • $\begingroup$ Also known as the method of differences. en.wikipedia.org/wiki/Telescoping_series $\endgroup$
    – George
    Feb 12 '16 at 21:26
  • $\begingroup$ I have just checked this online. I don't think I have seen this before. Can u please provide a brief description of the approach ? $\endgroup$
    – Room
    Feb 12 '16 at 21:28
  • $\begingroup$ Like you said write out the terms (I would write them down the page). Then you will see cancellation. Like on the wikipedia page: $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1-\frac{1}{n+1}$ - Notice the $\frac{1}{2}$'s etc cancel out $\endgroup$
    – George
    Feb 12 '16 at 21:30
  • $\begingroup$ alright thanks a lot ! $\endgroup$
    – Room
    Feb 12 '16 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.