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

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Answer 1 · 2016-12-27T14:04:37

#sum_(n=1)^oo (-1)^(n+1)/sqrt(n)# is convergent but not absolutely convergent.

To determine if:

#sum_(n=1)^oo (-1)^(n+1)/sqrt(n)#

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

#sum_(n=1)^oo (-1)^(n)a_n#

to converge, is that the succession #{a_n}# is decreasing and convergent to zero. In our case:

(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 #n#:

#1/sqrt(n) >= 1/n#

and

#sum_(n=1)^oo 1/n =oo#