I'm stuck with this problem from "An Introduction to Analysis by James Kirkwood" on page 48 exercise 24.

Let $\{x_n\}$ be a sequence of real positive numbers such that $\displaystyle\lim_{n\to \infty} \frac{x_{n+1}}{x_n}=L$ exists. Prove that if $L<1$ then $\{x_n\}$ converges to $0$.

I'm thinking (I got it from this similar question)

By definiton of limit I have that for any $\epsilon>0$ in particular for $0<\epsilon<1$ such that $L+\epsilon < 1$, exists $N_{\epsilon}\in \mathbb{N}$ such that $n>N_{\epsilon}$ implies that, $$|\frac{x_{n+1}}{x_n}-L|<\epsilon \Rightarrow \frac{x_{n+1}}{x_n}<L+\epsilon \Rightarrow x_{n+1}<(L+\epsilon)x_n<(L+\epsilon)^{n}x_n \\ \text{This will always happen even as $n \rightarrow \infty$, as long as $n>N_{\epsilon}$} $$

Since $\lim_{ N_{\epsilon}\to\infty }r^{N_{\epsilon}}=0$, when $0<r<1$, and because $0<(L+\epsilon)<1$ we must have that $x_{x+1}<0$ however because $x_n$ is a sequence of real positive numbers we have that $x_{n+1}$ must also be always greater than $0$. So $\{x_{n+1}\}$ converges to $0$, and therefore $\{x_n\}$ converges to $0$.

But I'm not sure about this thinking, since I think a possible contradiction might arise with $L$ being zero or something. Your thoughts?

  • $\begingroup$ What matters is that you get $0 < L+\varepsilon < 1$ to make this work. $\varepsilon$ is positive by assumption, so even if $L = 0$ you can restrict to the case $0 < \varepsilon < 1$ (as you actually do). $\endgroup$ – Gibbs Nov 7 '17 at 9:48
  • $\begingroup$ Your last inequality should not be $(L+\epsilon)x_n < (L+\epsilon)^nx_n$ but rather $(L+\epsilon)x_n < (L+\epsilon)^{n+1-N_\epsilon}x_{N_\epsilon}$. The limit that you are then looking at is $\lim_{n \rightarrow \infty} (L+\epsilon)^{n+1-N_\epsilon} = 0$. You cannot make $N_\epsilon$ go to infinity as it is simply a threshold after which your initial condition is verified. $\endgroup$ – M. P. Nov 7 '17 at 9:55
  • $\begingroup$ @M.P. I see what you mean. Thank you very much! $\endgroup$ – César Rosendo Nov 7 '17 at 10:09
  • 1
    $\begingroup$ Note that you can prove something stronger: the series $\sum_{n=0}^\infty x_n$ converges absolutely (see the ratio test en.wikipedia.org/wiki/Ratio_test). From the convergence of such series you can also deduce that $x_n\to 0$. Of course, this is not necessary to prove your claim, but still interesting to note, in my opinion. $\endgroup$ – Alberto Debernardi Nov 7 '17 at 11:01


$x_{n+1}/x_n \le L<1, x_n >0.$

This implies:

$\star)$ $x_n \lt x_0 L^n.$

Proof by induction:

$n=1$ : $x_1/x_0 \le L.$

Hypothesis: $\star)$.


$ x_n \le x_0 L^n;$

$ x_{n+1}(1/L) \le $

$x_{n+1}(x_n/x_{n+1}) \le x_0 L^n;$

$x_{n+1} \le x_0 L^{n+1}.$

$0 \le \lim_{n \rightarrow \infty} x_n \le $

$x_0 \lim_{n \rightarrow \infty} L^n = 0$.


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