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

##### 1 Answer

Dec 27, 2016

#### Explanation:

To determine if:

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

to converge, is that the succession

(i)

#lim_n 1/sqrt(n) = 0#

(ii)#a_(n+1)/a_n = (1/sqrt(n+1))/(1/sqrt(n))=sqrt(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

and