How do you determine the convergence or divergence of #Sigma ((-1)^(n+1)n)/(2n-1)# from #[1,oo)#?

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Answer 1 · 2016-12-23T09:28:23

The series:
#sum_(n=1)^oo (-1)^(n+1)n/(2n-1)#
is divergent.

You can determine whether an alternating series converges using Leibniz' criteria, which states that:

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

converges if:

(i) #a_n>a_(n+1)#
(ii) #lim_n a_n = 0#

As the general term of the series above can be expressed as:

#a_n = -n/(2n-1)#

We can quickly see that:

#lim_n a_n = lim_n-n/(2n-1)= lim_n-1/(2-1/n)=-1/2#

so that condition (ii) is not met and the series is divergent.