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.

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