Show that the sequence is absolutely convergent

Question

Assume that $\mid\frac{a_{n+1}}{a_n}\mid$ ≤ $\frac{n^2}{(n+1)^2}$ for n ∈ N. Show that the series $\sum$an is absolutely convergent.

By $\mid\frac{a_{n+1}}{a_n}\mid$ ≤ $\frac{n^2}{(n+1)^2}$ Can I assume $\mid a_n \mid$ ≤ $\frac{1}{n^2}$ ? If not how to continue please help.

Answer 1

No, you cannot assume that.

You have$$\require{cancel}|a_2|\leqslant\frac{1^2}{2^2}|a_1|\quad\text{and}\quad|a_3|\leqslant\frac{2^2}{3^2}|a_2|\leqslant\frac{\cancel{2^2}}{3^2}\frac{1^2}{\cancel{2^2}}|a_1|.$$Then you have$$|a_4|\leqslant\frac{3^2}{4^2}|a_3|\leqslant\frac{\cancel{3^2}}{4^2}\frac{1^2}{\cancel{3^2}}|a_1|$$and so on. More generally,$$(\forall n\in\Bbb N):|a_n|\leqslant\frac{|a_1|}{n^2}$$So, apply the comparison test, comparing your series to $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$.