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?
Best Answer
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, $\lim_nx_n=0$.