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Investigate the convergence or divergence properties of $\sum_{n=1}^{\infty}a_n$, where $a_n = \sqrt{n+1} - \sqrt{n}$.

I multiplied by its complex conjugate and resulted in $\frac{1}{\sqrt{n+1} + \sqrt{n}}.$ Then I tried applying the ratio test but that was inconclusive. I don't know how to apply the root test.

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Neither the root nor the ratio test will work. You do have this $$ \sqrt{n+1} - \sqrt{n} = {1\over \sqrt{n+1} + \sqrt{n} } \sim {1\over 2\sqrt{n}}. $$ How does $\sum_n 1/\sqrt{n}$ behave?

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Use telescoping to get \begin{align} \sum_{n=1}^N a_n & = \sum_{n=1}^N (\sqrt{n+1} - \sqrt{n})\\ & = (\sqrt{N+1} - \sqrt{N}) + (\sqrt{N} - \sqrt{N-1}) + \cdots + (\sqrt{3} - \sqrt2) + (\sqrt2- \sqrt1)\\ & = \sqrt{N+1} - 1 \end{align} Now conclude what you want.

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  • $\begingroup$ Nice direct solution. Mine was a little more of an arabesque but it works too. $\endgroup$ – ncmathsadist Dec 1 '13 at 0:19
  • $\begingroup$ I can't conclude from this, it just gives me a ratio test of $1$ again. $\endgroup$ – Don Larynx Dec 1 '13 at 0:32
  • $\begingroup$ @DonLarynx What is the definition of convergence of a series? Here we are looking at the sum of the first $N$ terms. $\endgroup$ – user17762 Dec 1 '13 at 0:32

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