How do you use the Nth term test on the infinite series #sum_(n=1)^oo(n(n+2))/(n+3)^2# ?

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Answer 1 · 2014-10-10T01:52:25

Let #a_n={n(n+2)}/{(n+3)^2}={n^2+2n}/{n^2+6n+9}#.

Since

#lim_{n to infty}a_n =lim_{n to infty}{n^2+2n}/{n^2+6n+9}#

by dividing the numerator and the denominator by #n^2#,

#=lim_{n to infty}{1+2/n}/{1+6/n+9/n^2}={1+0}/{1+0+0}=1 ne 0#,

the series diverges by Nth Term (Divergence) Test.

I hope that this was helpful.