convergent or divergent series?

**WartonMorton** · May 13th 2010, 01:28 AM

[sqrt(k) - sqrt(k-1)]^k

**tonio** · May 13th 2010, 01:54 AM

Originally Posted by WartonMorton

[sqrt(k) - sqrt(k-1)]^k

$\sqrt{k}-\sqrt{k-1}=\frac{1}{\sqrt{k}+\sqrt{k+1}}$ , and now raise all to the k-th power...

Tonio

**Failure** · May 13th 2010, 01:56 AM

Originally Posted by WartonMorton

[sqrt(k) - sqrt(k-1)]^k

Let's see

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

so, yes, that series converges.

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