Prove $\lim_{n}\sup{x_{n+1}\over x_n} < 1$ implies $\lim_n x_n = 0$

Question

Let $\{x_n\}$ be a sequence $x_n \ne 0$ such that: $$ \lim_{n\to\infty}\sup{x_{n+1}\over x_n} < 1 $$ Prove: $$ \lim_{n\to\infty}x_n = 0 $$

I would like to verify the below.

Let: $$ y_n = \frac{x_{n+1}}{x_n} $$ If $y_n$ converges, then: $$ \exists\lim_{n\to\infty}\sup y_n = L \implies \exists \lim_{n\to\infty}\sup |y_n| = |L| $$

Since $L < 1$, then $|L| \in [0, 1)$. Also: $$ \lim_{n\to\infty}y_n \le \lim_{n\to\infty}\sup y_n < 1 $$ Thus: $$ 0 \le \lim_{n\to\infty}|y_n| \le \lim_{n\to\infty}\sup |y_n| < 1 $$ Therefore: $$ 0 \le \lim_{n\to\infty}\left|\frac{x_{n+1}}{x_n}\right| < 1 $$

Which means $\exists N\in\Bbb N$ such that for $n > N$ the sequence of absolute values starts to strictly decrease. Thus: $$ \exists N\in\Bbb N : \forall n > N : |x_{n+1}| < |x_n| $$ But $|x_n| \ge 0$, then by monotone convergence theorem: $$ \lim_{n\to\infty} \left|\frac{x_{n+1}}{x_n}\right| \in [0, 1) \implies \lim_{n\to\infty}|x_n| = 0 \implies \lim_{n\to\infty}x_n = 0 $$

Could someone please verify the proof above?

Answer 1

The statement is wrong, a counter-example is $x_n = (-2)^n$ with $\limsup_{n\to\infty}{x_{n+1}\over x_n} = -2 < 1$.

There are several flaws in the proof:

• $y_n = \frac{x_{n+1}}{x_n}$ is not necessarily convergent.
• $\limsup y_n = L$ does not imply $\limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
• $L < 1$ does not imply $|L| \in [0, 1)$, a counter-example is $L=2$.
• If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.

You'll have to require that $\limsup_{n\to\infty} \left|x_{n+1}\over x_n\right| < 1$ instead. Then (as José already pointed out), $$ 0 \le |x_n| \le c^{n-n_0} |x_{n_0}| $$ for some $c \in [0, 1)$ and $n \ge n_0$, and $x_n \to 0 $ follows.

Answer 2

Your proof is wrong because you only proved that, if $n$ is large enough, then $\lvert x_{n+1}\rvert<\lvert x_n\rvert$ and that is not enough to deduce that $\lim_{n\to\infty}x_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.

Since $\limsup_n\left\lvert\frac{x_{n+1}}{x_n}\right\rvert<1$, if $c\in\left(\limsup_n\left\lvert\frac{x_{n+1}}{x_n}\right\rvert,1\right)$, then $\left\lvert\frac{x_{n+1}}{x_n}\right\rvert<c$, if $n\geqslant N$, for some $N\in\mathbb N$. So, $\lvert x_{N+1}\rvert\leqslant c\lvert x_n\rvert$, $\lvert x_{N+1}\rvert\leqslant c\lvert x_N\rvert$, $\lvert x_{N+2}\rvert\leqslant c^2\lvert x_N\rvert$, and so on. Therefore, by the squeeze theorem, $\lim_nx_n=0$.

Answer 3

The one weak step is when you go from the limit of the ratio $\left|\frac{x_{n+1}}{x_n}\right| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.

That step is true, but at the level of a proof, I think you need to justify why it must hold.