Trying to prove the Raabe-Duhamel test for divergance Let $a_n$ be a positive sequence.
If exists an $R\le 1$ and $N\in\mathbb{N}$, that for every $n>N$:
$\frac{a_{n+1}}{a_n}\ge 1-\frac{R}{n}$
so $\sum_{n=1}^\infty a_n$ diverges.
I proved that if $(na_n)$ is a positive monotone increasing sequence then $\sum_{n=1}^\infty a_n$ is divergant.
 A: $$a_{N+2}\geq a_{N+1}\left(1-\frac{R}{N+1}\right)$$
$$a_{N+3}\geq a_{N+2}\left(1-\frac{R}{N+2}\right)\geq a_{N+1}\left(1-\frac{R}{N+1}\right)\left(1-\frac{R}{N+2}\right)$$
And by induction, for $p\geq 1$,
$$a_{N+p+1}\geq a_{N+1}\prod_{k=N+1}^{N+p}\left(1-\frac{R}{k}\right)$$
And since $R\leq1$, you also have $1-\dfrac Rk\geq1-\dfrac1k$, hence
$$a_{N+p+1}\geq a_{N+1}\prod_{k=N+1}^{N+p}\left(1-\frac{1}{k}\right)=a_{N+1}\prod_{k=N+1}^{N+p}\left(\frac{k-1}{k}\right)=a_{N+1}\frac{N}{N+p}$$
And when $p\to\infty$, $a_{N+p+1}\sim\frac{\alpha}{p}$ with $\alpha>0$, so the series $\sum_na_n$ is divergent.
A: I haven't learned $\Pi$ yet, so I found a different solution:
First of all, we can see that $a_n\neq 0$ for every $n>N$ (because then $\frac{a_{n+1}}{a_n}$ is not defined).
$\frac{a_{n+1}}{a_n}\geq 1-\frac{R}{n}$
$n\cdot a_{n+1}\geq n\cdot a_n -R\cdot a_n$
$1\geq R$ so:
$n\cdot a_{n+1}\geq n\cdot a_n -R\cdot a_n\geq n\cdot a_n-a_n$
$n\cdot a_{n+1}\geq (n-1)\cdot a_n$
So the sequence $((n-1)a_n)_{n=1}^\infty$ is monotone increasing for every $n>N$. I'll prove that $\sum_{n=N}^\infty a_n$ diverges and finish.
For every $n>N$ the sequence is monotone increasing and greater than zero, so for every $n>N$:
$(n-1)\cdot a_n>(N-1)a_N>a_N$
$a_n >\frac{a_N}{n-1}$
$\sum_{n=1}^\infty a_n >\sum_{n=1}^\infty \frac{a_N}{n-1}$
I'll prove that $\sum_{n=1}^\infty \frac{a_N}{n-1}$ diverges and finish from the direct comparison test.
I'll use the limit comparison test with $\sum_{n=1}^\infty \frac{1}{n}$:
$lim_{n\to\infty} \frac{\frac{a_N}{n}}{\frac{1}{n-1}}=\frac{a_N\cdot (n-1)}{n}=a_N$
$\infty>a_N>0$ so the diverges from the limit comparison test.
