$\sum a_n = \sum{(\sqrt{n+1} - \sqrt{n})}$. I want to show that this diverges. I think the only way I can do it, is by using the comparison test, and finding a series less than it that diverges, but I can't think of a series that is used often/ easy to prove divergence for that is less than this one for all $n$. I was origionally thinking something like $\frac{1}{\sqrt{n}}$ or $\frac{1}{n}$, but those don't work
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3Do you notice anything special about the partial sum? $\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\cdots +\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1$? – hrkrshnn Nov 21 '14 at 14:24
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1If really you must do this by comparison, use that $$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}},$$ hence $$\frac1{2\sqrt{n+1}}\leqslant\sqrt{n+1}-\sqrt{n}\leqslant\frac1{2\sqrt{n}}.$$ – Did Nov 21 '14 at 14:25
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Observe that the sum telescopes $$\begin{align*}\sum_{n=1}^{N}\left(\sqrt{n+1}-\sqrt{n}\right)&=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\ldots+\sqrt{N+1}-\sqrt{N}=\\&=-1+\left(\sqrt{2}-\sqrt{2}\right)+\left(\sqrt{3}-\sqrt{3}\right)+\ldots+\left(\sqrt{N}-\sqrt{N}\right)+\sqrt{N+1}=\\\\&=\sqrt{N+1}-1 \quad \to \infty \end{align*}$$ as $N \to \infty$.
Jimmy R.
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Hint: Expand the first few terms, and derive a closed form for the sum.
