# $(x_{n})$ is a real sequence with $x_{n}\ne0$. Assume $\lim\frac{x_{n+1}}{x_{n}}=\ell$. Show $|\ell|<1\implies\lim{x_{n}}=0$.

The book's (A Problem Book in Real Analysis by Aksoy and Khamsi) solution goes like this:

$\lim\frac{x_{n+1}}{x_{n}}=\ell\implies\lim\left|\frac{x_{n+1}}{x_{n}}\right|=|\ell|$. So given $\varepsilon=\frac{1-|\ell|}{2}$, there exists $N_{0}\ge1$ such that for any $n\ge N_{0}$ we have $$\left| \frac{|x_{n+1}|}{x_{n}}-|\ell|\right|<\varepsilon\\\implies|\ell|-\varepsilon<\frac{|x_{n+1}|}{|x_{n}|}<|\ell|+\varepsilon$$ Then by definition of $\varepsilon$ we have $$\frac{|x_{n+1}|}{|x_{n}|}<\frac{|\ell|+1}{2}<1 \ \ \ \ \ \ \ (*)$$ For any $n\ge N_{0}$ $$|x_{n+1}|<\left(\frac{|\ell|+1}{2}\right)^{n-N_{0}+1}|x_{N_{0}}| \ \ \ \ \ \ \ (**)$$

Then $$\lim\left(\frac{|\ell|+1}{2}\right)^{n-N_{0}+1}=0\implies\lim|x_{n}|=0\implies\lim x_{n}=0$$

What I don't get:

1) How did they come up with $\varepsilon=\frac{1-|\ell|}{2}$? By $(*)$ it looks like they wanted something that when added to $|\ell|$ is $<1$, but I don't know why.

2) By multiplying $(*)$ by $|x_{n}|$ I see you can get something that looks like $(**)$, but how do you know $\left(\frac{|\ell|+1}{2}\right)^{n-N_{0}+1}|x_{N_{0}}|$ won't be smaller than $|x_{n+1}|$? How they did they know to raise $\frac{|\ell|+1}{2}$ to the power $n-N_{0}+1$?

3) Why does $0\le |x_{n+1}|<\text{sequence that converges to 0}$ allow us to conclude $\lim|x_{n}|=0$? How did they even know $\lim|{x_{n}}|$ exists?

Thanks in advance for any help.

• 3) suppose $|x_{n+1}|\leq a_n$ and $a_n\to0$. Then for any $\epsilon>0$ we can find $N$ with $n\geq N\implies ||x_n|-0|=|x_n|\leq a_{n-1}\leq\epsilon$. This states exactly that $|x_n|\to0$. Aug 18, 2017 at 7:35

(1) Given $|\ell|<1$, you want to find $\ell'$ so that $|\ell|<\ell'<1$ and

$$\frac{|x_{n+1}|}{|x_n|}<\ell'$$

when $n$ is large. One such $\ell'$ is to take $\ell'$ as the midpoint of $|\ell|$ and $1$. That is $|\ell'| = \frac{|\ell|+1}{2}$ and so

$$\ell' - |\ell| = \frac{|\ell|+1}{2} - |\ell| = \frac{1-|\ell|}{2}$$

is the epsilon they choose. It doesn't have to be this $\ell'$ (and $\epsilon$). Any $\ell'$ so that $|\ell|<\ell'<1$ would do the job.

(2) So $\frac{|x_{n+1}|}{|x_n|}<\frac{1+|\ell|}{2}$ for all $n\ge N_0$. In particular multiplying the inequalities for $n=N_0$ and $n=N_0+1$ gives

$$\frac{|x_{N_0+2}|}{|x_{N_0}|}= \frac{|x_{N_0+2}|}{x_{N_0+1}|}\frac{|x_{N_0+1}|}{|x_{N_0}|}< \left( \frac{1+|\ell|}{2}\right)^2.$$

Which proves the inequality for $n = N_0 +1$. In general you can multiply $n$'s inequalities to obtain (**). Of course the rigorous way is to do induction.

(3) This is sandwich's theorem (aka squeeze theorem), where the existence of limit is part of the conclusion.