# Proving that $\sum_ {n=1}^{\infty}{\frac{(-1)^n}{\sqrt{n}}}$ converges

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. –  André Nicolas May 29 '12 at 5:58

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.