How do I determine if the alternating series #sum_(n=1)^oo(-1)^n/sqrt(3n+1)# is convergent?

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Answer 1 · 2014-10-18T11:08:55

Alternating Series Test

An alternating series #sum_{n=1}^infty(-1)^n b_n#, #b_n ge 0# converges if both of the following conditions hold.

#{(b_n ge b_{n+1} " for all " n ge N),(lim_{n to infty}b_n=0):}#

Let us look at the posted alternating series.

In this series, #b_n=1/sqrt{3n+1}#.

#b_n=1/sqrt{3n+1} ge 1/sqrt{3(n+1)+1}=b_{n+1}# for all #n ge 1#.

and

#lim_{n to infty}b_n=lim_{n to infty}1/sqrt{3n+1}=1/infty=0#

Hence, we conclude that the series converges by Alternating Series Test.

I hope that this was helpful.