Proving a contrapositive of the ratio test for sequences.

Question

Let $(a_n)$ be a sequence s.t. $a_n\rightarrow L\in\mathbb{R}\setminus{\{0\}}$, let $\forall n\in\mathbb{N},a_n>0$. We wish to show that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$. (From definition). My go:

By assumption, for any $\epsilon>0$ there is some $n_0$ for which $L-\epsilon<a_n<L+\epsilon$. Now, divide through by $a_{n+1}$ to obtain $$\frac{L-\epsilon}{a_{n+1}}<\frac{a_n}{a_{n+1}}<\frac{L+\epsilon}{a_{n+1}}$$ now as $a_{n+1}$ is a subsequence of $a_n$ it follows that $a_{n+1}\rightarrow L$, hence $$\frac{L-\epsilon}{L}<\frac{a_n}{a_{n+1}}<\frac{L+\epsilon}{L}$$ now further simplify to $$|\frac{a_{n}}{a_{n+1}}-1|<\epsilon/L$$ which holds for all $\epsilon>0$ and thus $\frac{a_{n}}{a_{n+1}}\rightarrow 1$. Would that be a correct approach?

Answer 1

First observe that $L > 0$, and for any $\epsilon > 0$, $\exists N \ge 1$ such that: if $n \ge N$, then : $|a_n - L| < \dfrac{L}{2} \implies a_n > \dfrac{L}{2}$, and $|a_{n+1} - L| < \dfrac{L\epsilon}{4}$, $|a_n - L| < \dfrac{L\epsilon}{4}$. Thus: if $n \ge N$, then: $\left|\dfrac{a_{n+1}}{a_n} - 1\right|= \left|\dfrac{(a_{n+1} - L) + (L - a_n)}{a_n}\right|\le \dfrac{1}{a_n}\left(|a_{n+1}-L|+|a_n - L|\right)< \dfrac{2}{L}\cdot \left(\dfrac{L\epsilon}{2}\right)=\epsilon$. Proving the claim.

Answer 2

Guide:

Given $\epsilon >0$, we know that there is $N_1>0$ such that $n > N_1$, then $|a_n -L| < \frac{L}{2}$. Hence if $n > N_1$, we have $a_n > \frac{L}2$, and hence $\frac{1}{a_n}< \frac2L.$

\begin{align} \left| \frac{a_{n+1}}{a_n} -1\right| &= \frac{|a_{n+1}-a_n|}{|a_n|} \\ &\le \frac{|a_{n+1}-L|+|a_n-L|}{|a_n|} \\ &= \frac{2}{L}\left(|a_{n+1}-L|+|a_n-L| \right) \end{align}

Hopefully you can see how to pick $N$ from here.