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?

  • 3
    $\begingroup$ Well, $\frac{1}{\sqrt{n+1}}\lt \frac{1}{\sqrt{n}}$, but that happens to be what you need. $\endgroup$ May 29, 2012 at 5:58

1 Answer 1


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.


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