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Need help finding a minorant to $(\sqrt{k+1} - \sqrt{k})$ which allows me to show that the series $\sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})$ is divergent.

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5 Answers

Do you need a minorant? Just consider the partial sums. By telescoping you see that $\sum\limits_{k=1}^n \sqrt{k+1} - \sqrt k = \sqrt{n+1} - 1$, which diverges.

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You should observe that your series telescopes, i.e.: $$\sum_{k=0}^n (\sqrt{k+1} - \sqrt{k}) = (\sqrt{1} -\sqrt{0}) + (\sqrt{2} -\sqrt{1}) +\cdots + (\sqrt{n}-\sqrt{n-1}) +(\sqrt{n+1}-\sqrt{n}) = \sqrt{n+1}-1\; ,$$ and therefore: $$\sum_{k=0}^\infty (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sum_{k=0}^n (\sqrt{k+1} - \sqrt{k}) = \lim_{n\to \infty} \sqrt{n+1}-1 = \infty\; .$$

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Since $$\sum_{k=1}^n (\sqrt{k+1} - \sqrt{k}) =\sum_{k=1}^n ((\sqrt{k+1} - \sqrt{k})\frac{\sqrt{k+1} + \sqrt{k}}{\sqrt{k+1} + \sqrt{k}}) $$ $$ =\sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \geq 2\sum_{k=1}^n \frac{1}{\sqrt{k+1}} \geq 2\sum_{k=1}^n \frac{1}{k+1}, $$ then the series does not converge, but the telescope argument is much simpler.

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If you do want a minorant, then you can use $$ \sqrt{k+1}-\sqrt k = \int_k^{k+1} \frac1{2\sqrt x} dx > \int_k^{k+1} \frac1{2\sqrt {k+1}} dx = \frac1{2\sqrt {k+1}}. $$

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Note that $$ \sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}>\frac{1}{2\sqrt{k+1}} $$

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This might be similar to miracle173's answer, but I have trouble parsing that answer. –  robjohn Mar 26 '12 at 19:33
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