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Prove that the following series converges:

$\sum_ {n=1}^{\infty}{\frac{(-1)^n}{\sqrt{n}}} $

$\frac{1}{\sqrt{n+1}} > \frac{1}{\sqrt{n}}$ $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$ So, the alternating series converges.

Is it right to my procedure?

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    Well, $\frac{1}{\sqrt{n+1}}\lt \frac{1}{\sqrt{n}}$, but that happens to be what you need.2012-05-29

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To use Alternating series test for series $\sum(-1)^n a_n$ you should have

  1. $a_n\geq 0$

  2. $a_n\geq a_{n+1}$

  3. $\lim a_n = 0$

In your approach you wrote that $a_{n+1} = \frac{1}{\sqrt{n+1}}>\frac{1}{\sqrt n} = a_{n}$ which is wrong and not what you need. Fix it and your solution will be correct.