# Prove that the series $\sum_{k=0}^\infty (-1)^k\ \left(\sqrt{k+1}-\sqrt{k}\right)$ converges but not absolutely.

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?

• What does k run from? $1$ to $\infty$? – George Feb 12 '16 at 21:12
• the summation runs from 0 to ∞ @George – Room Feb 12 '16 at 21:16
• The series diverges. Perhaps you should check the source to see if you have copied it correctly. – John Bentin Feb 12 '16 at 21:32
• 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. – Steven Stadnicki Feb 12 '16 at 21:48

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 $N\to \infty$.
• 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 – George Feb 12 '16 at 21:30