Prove Raabe's test for the convergence of a series

Consider the series $\sum a_n$ where $a_n > 0$ for all $n$ . Raabe’s test for convergence states that if there exists an integer $N \geq 1$ and a constant $r > 0$ such that

$\frac{a_{n+1}}{a_n} \leq 1 - \frac{1}{n} - \frac{r}{n} \qquad \text{for all } n \geq N$

then the series $\sum a_n$ is convergent. The series is divergent if

$\frac{a_{n+1}}{a_n} \geq 1 - \frac{1}{n} \qquad \text{for all } n \geq N.$

Proof. Assume there is an $r > 0$ and an $N \geq 1$ such that

$\frac{a_{n+1}}{a_n} \leq 1 - \frac{1}{n} - \frac{r}{n} \qquad \text{for all } n \geq N.$

From the previous exercise (Section 10.16, Exercise #15) we know that if there exists an integer $N$ and an $r >0$ such that

$0 < r \leq b_n - \frac{b_{n+1}}{a_{n+1}}{a_n} \qquad \text{for all } n \geq N$

then $\sum a_n$ converges. Taking $b_n = n$ , we have

$\begin{align*} && \frac{a_{n+1}}{a_n} &\leq 1 - \frac{1}{n} -\frac{r}{n} \\[9pt] \implies && \frac{n(a_{n+1})}{a_n} &\leq n - 1 - r \\[9pt] \implies && \frac{n(a_{n+1})}{a_n} + 1 - n &\leq -r \\[9pt] \implies && r &\leq n - n \left( \frac{a_{n+1}}{a_n} \right) - 1 \\[9pt] \implies && r &\leq n - \left( \frac{na_{n+1}}{a_n} + 1 \right) \\[9pt] \implies && r &\leq n - \left( \frac{na_{n+1}}{a_n} + \frac{a_{n+1}}{a_n} \right) &(\frac{a_{n+1}}{a_n} \leq 1 \text{ by assumption})\\[9pt] \implies && r &\leq n - \frac{(n+1)a_{n+1}}{a_n} \\[9pt] \implies && r &\leq b_n - \frac{b_{n+1}a_{n+1}}{a_n}. \end{align*}$