[sqrt(k) - sqrt(k-1)]^k
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Originally Posted by WartonMorton [sqrt(k) - sqrt(k-1)]^k $\displaystyle \sqrt{k}-\sqrt{k-1}=\frac{1}{\sqrt{k}+\sqrt{k+1}}$ , and now raise all to the k-th power... Tonio
Originally Posted by WartonMorton [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.
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