- #1

The-Mad-Lisper

- 12

- 1

## Homework Statement

[itex]\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}[/itex]

## Homework Equations

[itex]S=\sum\limits_{n=1}^{\infty}a_n[/itex] (1)

[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent[/itex] (2)

[itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent[/itex] (3)

[itex]\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent[/itex] (4)

## The Attempt at a Solution

I suspect that the given series is divergent, thus the series would not be convergent. Thus I take the contrapositive of statement 4 which comes out to be [itex]\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent[/itex].

Let [itex]a_n=\frac{n-1}{(n+2)(n+3)}[/itex] and [itex]b_n = \frac{1}{n}[/itex]. The statement [itex]\mid b_n \mid\ \leq\ \mid a_n[/itex] is never true in the domain [itex][1,\infty)[/itex], so we cannot say that [itex]a_n[/itex] is divergent. Any suggestions?