# Math Help - convergent or divergent series?

1. ## convergent or divergent series?

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

2. 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

3. 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.