# How do you determine if the series the converges conditionally, absolutely or diverges given Sigma (-1)^(n+1)/sqrtn from [1,oo)?

Dec 27, 2016

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \sqrt{n}$ is convergent but not absolutely convergent.

#### Explanation:

To determine if:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \sqrt{n}$

we can use use Leibniz' test, which states that a sufficient condition for an alternating series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$

to converge, is that the succession $\left\{{a}_{n}\right\}$ is decreasing and convergent to zero. In our case:

(i) ${\lim}_{n} \frac{1}{\sqrt{n}} = 0$
(ii) ${a}_{n + 1} / {a}_{n} = \frac{\frac{1}{\sqrt{n + 1}}}{\frac{1}{\sqrt{n}}} = \sqrt{\frac{n}{n + 1}} < 1$

so the both conditions are met and the series is convergent.

We can also conclude that the series is not absolutely convergent, by direct comparison, since for any $n$:

$\frac{1}{\sqrt{n}} \ge \frac{1}{n}$

and

${\sum}_{n = 1}^{\infty} \frac{1}{n} = \infty$