How do you test the alternating series #Sigma (-1)^(n+1)/sqrtn# from n is #[1,oo)# for convergence?

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Answer 1 · 2017-05-16T06:25:20

A sufficient condition for an alternating series #sum (-1)^na_n# to converge is that:

(1) #lim a_n= 0#

(2) #a_(n+1) <= a_n#

The series:

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

is then convergent, since:

#lim 1/sqrt(n) = 0#

and

#1/sqrt(n+1) < 1/sqrt(n)#