# Proof for the Ratio Test

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. $$\newcommand{\absval}{\left\lvert #1 \right\rvert}$$

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.

• I can't seem to find anything wrong with your proof, It's pretty much how I would have written it. Jan 1, 2021 at 21:59

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.