Question: Suppose that $\{x_n\}$ is a sequence and suppose that for some $x\in \mathbb{R}$, $$ L=\lim_{n \to \infty}\frac{ x_{n+1} - x}{x_n - x} $$ exists and $L<1$. Prove that the sequence $\{x_n\}$ converges to $x$.

What I know: A sequence $\{x_n\}$ is said to converge to a number $x \in \mathbb{R}$, if for every $\epsilon>0$, there exists an $M \in \mathbb{N}$ such that $\left\vert\,x_{n} - x\,\right\vert < \epsilon$ for all $n\ge M$. The number $x$ is said to be the limit of ${x_n}$. So the $\lim_{n \to \infty} x_n = x$.

How do I even begin to write this proof?

  • $\begingroup$ Have you already studied series? $\endgroup$ – DonAntonio Dec 4 '13 at 20:55

Just prove it for $x=0$,because if we let $y_n=x_n-x$,then $y_n\to 0<=>x_n\to x$. Now it's easier to show that if $|\frac {y_{n+1}}{y_n}|\to L<1=>|y_n|\to 0$.

To make things easier. Suppose that $y_n> 0$for every $n$. Then $\frac {y_{n+1}}{y_n}\to L<=> \forall ε>0$ $\exists N:|\frac {y_{n+1}}{y_n}|\leq L<1$ ,$\forall n\geq N$.

So for $n\geq N$ we have that $y_{n+1}< y_n$(decreasing strictly). So?does it go to $0$?

  • $\begingroup$ It goes to 1 wouldn't it? $\endgroup$ – Cynthia Dec 4 '13 at 22:57
  • $\begingroup$ If you mean $y_n$,then it goes to $0$ and thus $x_n\to x$. $\endgroup$ – Haha Dec 4 '13 at 23:02
  • $\begingroup$ so it gets arbitrarily small but doesn't hit zero? How would I go about creating a proof that flows? $\endgroup$ – Cynthia Dec 4 '13 at 23:48
  • $\begingroup$ In general,we have that $|y_{n+1}|<L|y_n|$ for $n\geq N$ and thus,$|y_{(N+1)+(k+1)}|<L^k|y_N|\to 0$ for $k\to \infty$ because $L^k\to 0$ because $L<1$.And thus $|y_n|\to 0=>y_n\to 0$ $\endgroup$ – Haha Dec 5 '13 at 0:04

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