convergent or divergent series?

Jul 2009
555
298
Zürich
[sqrt(k) - sqrt(k-1)]^k
Let's see

\(\displaystyle 0 \leq \left[\sqrt{k}-\sqrt{k-1}\right]^k=\left[\frac{1}{\sqrt{k}+\sqrt{k-1}}\right]^k\leq \left[\frac{1}{2\sqrt{k-1}}\right]^k\leq \frac{1}{2^k}\), for \(\displaystyle k\geq 2\)

so, yes, that series converges.