I'd like to ask, if my below proof for the ratio test for the convergence of an infinite series is technically correct and rigorous.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
Given a series $\sum_{n=1}^{\infty}a_n$ with $a_n \ne 0$, the Ratio Test states that if $(a_n)$ satisfies
\begin{align*}
\lim \absval{\frac{a_{n+1}}{a_{n}}} = r < 1
\end{align*}
then the series converges absolutely.
(a) Let $r'$ satisfy $r < r' < 1$. Explain why there exists an $N$ such that $n \ge N$ implies that $\absval{a_{n+1}} \le \absval{a_n} r'$.
(b) Why does $\absval{a_N} \sum (r')^n$ converge?
(c) Now, show that $\sum \absval {a_n}$ converges, and conclude that $\sum a_n$ converges.
Proof.
(a) Since, $\lim (a_{n+1}/a_{n}) \to r$, given any $\epsilon > 0$, there exists $N \in \mathbf{N}$, such that :
\begin{align*}
\frac{\absval{a_{n+1}}}{\absval{a_{n}}} < r + \epsilon
\end{align*}
If we set $r + \epsilon = r'$, we have the desired inequality,
\begin{align*}
\absval{a_{n+1}} < r'\absval{a_n}
\end{align*}
for all $n \ge N$. Moreover, since the limit of the sequence of ratios is less than unity, the ratios themselves should also be less than unity (Using the fact that $a_n \le b_n \Longleftrightarrow \lim a_n \le \lim b_n$ ,the order limit theorem). Therefore, $\absval{a_{n+1}}/\absval{a_n} < 1$ for all $n \in \mathbf{N}$. Therefore, $r < r' < 1$.
(b) $\absval{a_N}\sum (r')^n$ is a geometric series. Since, the partial sums of this series, are monotone increasing, but bounded (by $a_{N}/(1-r')$), by the monotone convergence theorem, $\absval{a_N}\sum (r')^n$ is convergent.
(c) We have, $\absval{a_{n+1}} \le \absval{a_N}(r') \le \absval{a_N}(r')^n$ for all $n \ge N$. As $\absval{a_N}\sum (r')^n$ is convergent, by the Comparison test, $\sum \absval {a_n}$ is also convergent. By the Absolute value test, $\sum a_n$ is absolutely convergent.
Best Answer
The ideas are right here. To make it clearer for (a) I would write: Let $r' \in (r,1)$, then there exists $\epsilon > 0$ such that $r + \epsilon = r'.$ Then continue from there as you have finding the $N$ you need for the $\epsilon$ given. In your answer it read as if you were only considering one value of $r'$ although of course this isn't what you mean. You also don't need to explain why $r < 1$ here since it's given in the question (although yes, that is right).
For (b) just write that since $0 < r' < 1$ it's a geometric series. You don't need the other parts (even though it's correct).
The ideas in (c) look right but it's a little unclear. Let $N$ be from $(a)$ and note that
$$\sum_{n=1}^\infty |a_{N+n}| \leq \sum_{n=1}^\infty |a_N|r'^{n}.$$
Then continue as you have to explain why this converges. Note that we needed to shift our sum along by $N,$ so we aren't dealing with exactly the same series, so a sentence or two to explain how we then deduce that $\sum a_n$ indeed converges.