# Proof for Convergent of Series With Seq. Similar to 1/n

Tags:
1. Apr 6, 2016

1. The problem statement, all variables and given/known data
$\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}$

2. Relevant equations
$S=\sum\limits_{n=1}^{\infty}a_n$ (1)
$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent$ (2)
$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent$ (3)
$\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent$ (4)

3. 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 $\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent$.
Let $a_n=\frac{n-1}{(n+2)(n+3)}$ and $b_n = \frac{1}{n}$. The statement $\mid b_n \mid\ \leq\ \mid a_n$ is never true in the domain $[1,\infty)$, so we cannot say that $a_n$ is divergent. Any suggestions?

2. Apr 6, 2016

### Samy_A

That's a good idea, comparing with the harmonic series $\sum \frac{1}{n}$.

Hint: $\frac{n-1}{n+2}$ is close to 1 for large n.

3. Apr 7, 2016

### LCKurtz

@The-Mad-Lisper Have you studied the limit comparison test yet?

4. Apr 13, 2016

Only after the fact, I managed to modify Nicole Oresme's proof of the divergent nature of a harmonic series.

5. Apr 13, 2016

### Ray Vickson

The terms of your sum are all positive, so it is enough to give a lower bound series that diverges.

For $n > 6$ we have $n-1 > \frac{1}{2}(n+3)$ and $(n+2)(n+3) < (n+3)^2$, so
$$\sum_{n=6}^{\infty} \frac{n-1}{(n+2)(n+3)} > \sum_{n=6}^{\infty} \frac{1}{2} \frac{n+3}{(n+3)^2} = \frac{1}{2} \sum_{k=9}^{\infty} \frac{1}{k}$$