How do you determine the convergence or divergence of #Sigma ((-1)^(n+1)n^2)/(n^2+5)# from #[1,oo)#?

1 Answer
May 12, 2017

The non-alternating portion of the sequence that makes up the series is #n^2/(n^2+5)#.

Note that #lim_(nrarroo)n^2/(n^2+5)=1#.

This means that as #n# grows infinitely larger, the series will begin to resemble #sum_(n=1)^oo(-1)^n#, which diverges because it never "settles".

A powerful tool for testing divergence is the rule that if #suma_n# converges, then #lim_(nrarroo)a_n=0#. That is, as #n# gets larger, the terms of any convergent series will approach #0#.

Here, #lim_(nrarroo)a_n=1#, so the series is divergent.