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I want to know if the sum $$\sum_{n=1}^\infty(-1)^n\frac{\sqrt{n+1}-\sqrt{n}}{n}$$ converges or not. So I've tried the alternating series text:

  1. $\lim\limits_{n\rightarrow\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=0$ is clear.
  2. I also need $a_{n+1}\leq a_n$.

So my question is if there is an very easy way to show 2. ? (I've tried to calculate the inequation but I don't get a nice result.)

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Hint:

$$\frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{n(\sqrt{n+1}+\sqrt{n})}=\frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$

Clearly, $a_n$ is decreasing.

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    $\begingroup$ Which also shows that this sum absolutely converges. $\endgroup$ – Amihai Zivan Jan 1 '13 at 14:39
  • $\begingroup$ so I have $\frac{1}{(n+1)(\sqrt{n+2}+\sqrt{n+1})}\leq\frac{1}{n(\sqrt{n+1}+\sqrt{n})}$ but why does it help me? $\endgroup$ – user45389 Jan 1 '13 at 14:56
  • $\begingroup$ This shows that $a_n$ is decreasing $\endgroup$ – Amr Jan 1 '13 at 14:58
  • $\begingroup$ sry but I don't see it :( $\endgroup$ – user45389 Jan 1 '13 at 15:00
  • $\begingroup$ You wanted to show that $a_n\geq a_{n+1}$. Now you are done, because $a_n=\frac{1}{n(n^{0.5}+(n+1)^{0.5})}$ is decreasing $\endgroup$ – Amr Jan 1 '13 at 15:01
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The answer by @Amr is very nice and solves your problem. Here are some other approaches.

I. Let $a_n = (\sqrt{n+1}-\sqrt{n})/n$. Then $$\begin{eqnarray*} \frac{a_{n+1}}{a_n} &=& \frac{\sqrt{n+2}-\sqrt{n+1}}{n+1} \frac{n}{\sqrt{n+1}-\sqrt{n}} \\ &=& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1}) \\ &<& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n+2}+\sqrt{n+1}) \\ &=& \frac{n}{n+1}. \end{eqnarray*}$$

II. Since $$\frac{a_{n+1}}{a_n} \sim 1 - \frac{3}{2n} + \ldots \qquad (n\to\infty)$$ the series converges absolutely by Raabe's test.

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Another way is to define a function $f$ where $f(x)=\displaystyle\frac{\sqrt{x+1}-\sqrt{x}}{x}$ you get

$f'(x)=\dfrac{\sqrt{{x}^{2}+x}-x-2}{2{x}^{2}{\cdot}\sqrt{x+1}}$ which is strictly negative since $\sqrt{x^2+x}<x+2$ hence $f$ is decreasing and so is $a_n$.

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