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Let $x_n$ be a positive sequence such that the sequence $(\displaystyle\frac{x_{n+1}}{x_n})$ converges to $\lambda<1$. Show that $x_n$ converges to $0$.

Hint: Show that there exists $c,r$ such that $0<r<1$ and $0\leq x_n\leq cr^n$

So I managed to prove that for $n\geq R$, using the definition of convergence of a sequence, $x_n$ is an decreasing sequence. From there, I managed to get the following:

$$0\leq \frac{x_n}{x_R} \leq \frac{x_{n+1}}{x_n} < \lambda+\epsilon, \forall \epsilon > 0$$ $$\implies0\leq \frac{x_n}{x_R} \leq \lambda$$ $$\implies 0\leq x_n \leq x_R\lambda$$

So I feel the candidates for $c,r$ are $x_R,\lambda$, but from here I'm not sure how to get that $0\leq x_n\leq cr^n$ to conclude that $x_n$ converges to $0$.

Any help would be appreciated.

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  • $\begingroup$ So I'm unsure you can show that $0 \le x_n$ since $x_n = -\frac{1}{2^n}$ fits the bill (with $\lambda = \frac{1}{2}$), but every term is negative. $\endgroup$ – DanZimm Apr 6 '15 at 4:38
  • $\begingroup$ @DanZimm sorry, forgot to mention the sequence was positive. $\endgroup$ – Rono Apr 6 '15 at 4:46
  • $\begingroup$ No problem! I would check out mich's answer, seems to be exactly what I was thinking! $\endgroup$ – DanZimm Apr 6 '15 at 4:48
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I have no idea whether the result holds for sequences of alternating sign. But if $x_{n} >0$, the result probably holds. Since $x_{n} >0$ then $\lambda \geq 0$. for all $\epsilon>0$ there exists $N(\epsilon)$ such that for all n $\geq N(\epsilon)$, |$\frac{x_{n+1}}{x_{n}}-\lambda$| $<\epsilon$. Since $0\leq \lambda < 1$, we can find r such that $\lambda <r<1$. Let $\epsilon = r-\lambda$. Hence we have an integer K such that if $n \geq K$ , |$\frac{x_{n+1}}{x_{n}}-\lambda$| $<r-\lambda$, hence $0<\frac{x_{n+1}}{x_{n}} <r$. So $0<x_{n+1}<rx_{n}<r^{2}x_{n-1}< \cdots <x_{K}r^{n-K+1}=X_{K}r^{-K}r^{n+1}$ and the result follows by squeeze theorem, since $\lim\limits_{n \to \infty} r^{n}=0$,$(0<r<1$).

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  • $\begingroup$ Yeah, forgot to mention the sequence had to be positive. How do you get that $rx_n < r^2x_{n-1}$? That's what I'm not getting. $\endgroup$ – Rono Apr 6 '15 at 4:48
  • $\begingroup$ If $n \geq K$ then $x_{n+1} \leq rx_{n}$, and you keep applying the formula recursively, until you reach $x_{K}$. $\endgroup$ – mich95 Apr 6 '15 at 5:27
  • $\begingroup$ Cool, thanks for this! $\endgroup$ – Rono Apr 6 '15 at 6:20
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You know that all terms of the quotient above a certain $N$ are within $[0,r]$ with $0 < \lambda \le r <1$.

Then what happens to $\frac{x_{N+1}}{x_N} \times \ldots \times \frac{x_{n}}{n-1}$ ?

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