2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ I can't seem to find anything wrong with your proof, It's pretty much how I would have written it. $\endgroup$ – J.V.Gaiter Jan 1 at 21:59
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.