Derivation of refined Ratio test of convergence with Bernoulli inequallity I am asked to derive the "General Ratio Test" (by using the Bernoulli inequality) with is stated as follows:
$(a_j)_{j\in\mathbb{N}}$ is a sequence in $\mathbb{R}_{>0}$. Show that the series $\sum_j{a_j}$ converges in the case there is a $\theta \in \mathbb{R}$ with $\theta > 1 $ and an $N\in\mathbb{N}$, so that:
\begin{equation}
\frac{a_{j+1}}{a_j}\leq1-\frac{\theta}{j+1},\text{ for all }j\geq N
\end{equation}
However, my problem is that I must use the Bernoulli Inequality, which is as follows:
a) For $q=n/m$, with $n,m \in \mathbb{N}_{\geq 1}$ with $n\leq m$
\begin{equation}
(1+x)^q\leq 1+qx
\end{equation}
b) For $q\in\mathbb{Q}_{\geq 1}$
\begin{equation}
(1+x)^q\geq 1+qx
\end{equation}
My approach was the following:
\begin{equation}
\frac{a_j}{a_{j-1}}\leq 1-\frac{\theta}{j} \\
a_j \leq a_{j-1}\left(1-\frac{\theta}{j}\right)\leq a_{j-2}\left(1-\frac{\theta}{j}\right)^2\leq ... \leq a_N\left(1-\frac{\theta}{j}\right)^{j-N}
\end{equation}
With a variable $j'=j-N$ change, I get:
\begin{equation} 
\sum_{j=N}^n a_j \leq \sum_{j=N}^n a_N\left(1-\frac{\theta}{j}\right)^{j-N}=\sum_{j'=0}^{n-N} a_N \left(1-\frac{\theta}{j'+N}\right)^{j'}
\end{equation}
I thout I could use b) but apparently it wouldn't give me any insights into the convergence. I suppose my approach is wrong, or there is something I am missing.
I appreciate your answers.
 A: At this point your approach is incorrect since
$$\left(1-\frac{\theta}{j}\right)^{j-N} = \left(1-\frac{\theta}{j}\right)^{-N}\left(1-\frac{\theta}{j}\right)^{j} \underset{j \to \infty}\longrightarrow 1 \cdot e^{-\theta} \neq 0,$$
and the upper bound $\displaystyle\sum_{j=N}^n\left(1-\frac{\theta}{j}\right)^{j-N}$ diverges as $n \to \infty$.
Straightforward Proof
Suppose $a_j > 0$ for all $j$ and we have $\displaystyle\frac{a_{j+1}} {a_j} \leqslant 1 - \frac{\theta}{j+1}$ for all $j \geqslant N$, where $\theta> 1$.  It follows that
$$\frac{(j+1)a_{j+1}}{a_j} \leqslant j+1 - \theta = j - (\theta-1),$$
which implies that $(\theta-1) a_j \leqslant ja_j - (j+1)a_{j+1}.$
Summing over $j = N, \ldots, M$, we get
$$S_M= \sum_{j = N}^Ma_j = (\theta-1)^{-1}[Na_N - (M+1) a_{M+1}] < (\theta-1)^{-1}Na_N,$$
and the partial sums $S_M$ are increasing and bounded for all $M > N$ and, therefore, convergent as $M \to \infty$.
Therefore,
$$\sum_{j=1}^\infty a_j = \sum_{j=1}^{N-1} a_j + \sum_{j=N}^\infty a_j < +\infty$$
