Prove that $\displaystyle \sum_{k=1}^\infty (-1)^{k+1}(\sqrt{k+1}-\sqrt{k})$ is convergent. Is it absolutely convergent? If so prove it if not then find a rearrangement where the series sums to a number greater than $1$.
Theorem: Suppose $\{a_n\}$ is monotonically decreasing and $\lim_{n\rightarrow\infty}a_n=0$ then $\sum_{k=1}^\infty (-1)^{k+1}a_k$ converges.
My attempt: First note that $\sqrt{k+1}-\sqrt{k}=(\sqrt{k+1}-\sqrt{k})(\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k+1}+\sqrt{k}})=\frac{1}{\sqrt{k+1}+\sqrt{k}}$. Then $\lim_{k\rightarrow\infty}\frac{1}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\infty}=0$.
Now let $s_k=\sqrt{k+1}-\sqrt{k}$, so $s_{k+1}=\sqrt{k+2}-\sqrt{k+1}$. Then $$s_{k+1}-s_k=\sqrt{k+2}-\sqrt{k+1}-\sqrt{k+1}+\sqrt{k}$$ $$=\sqrt{k+2}-2\sqrt{k+1}+\sqrt{k}$$ $$<(k+2)-4(k+1)+k$$ $$=-2k-2<0$$ Thus, $\{s_k\}$ is monotonically decreasing, so by the theorem, $\displaystyle \sum_{k=1}^\infty (-1)^{k+1}(\sqrt{k+1}-\sqrt{k})$ converges.
Now for absolutely convergence, we have to check if $\displaystyle \sum_{k=1}^\infty |(-1)^{k+1}(\sqrt{k+1}-\sqrt{k})|$ converges. Notice that $$\displaystyle \sum_{k=1}^\infty |(-1)^{k+1}(\sqrt{k+1}-\sqrt{k})|=\displaystyle \sum_{k=1}^\infty (\sqrt{k+1}-\sqrt{k})$$.
Since this diverges, it is not absolutely convergent. The part that I am stuck on is the rearrangement part. Could I get a hint on how to rearrange the series to get a number greater than 1? Thanks.