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

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Answer 1 · 2017-01-03T15:16:50

#sum_(n=1)^oo (-1)^n n^2/(n^2+1)# does not converge

This is an alternating series, so the necessary condition for it to converge is that:

#lim a_n = 0#

#a_(n+1)/a_n < 1#

As:

#a_n = n^2/(n^2+1)#

we have:

#lim_n a_n = lim_n n^2/(n^2+1) = 1#

Therefore the series does not converge.