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Let $X=(x_n)$ be a sequence of strictly positive real numbers such that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$. Show that for some $r$ with $0<r<1$ and some $C>0$, then we have $0<x_n<Cr^n$ for all sufficently large $n\in \mathbb{N}$. Use this to show that $\text{lim}(x_n)=0$. Similarly, show that $\lim\left(\frac{x_{n+1}}{x_n}\right)>1$ is not bounded and hence not convergent.

Since I know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$ and that $0<r<1$, then I have $0<\lim\left(\frac{x_{n+1}}{x_n}\right)\le r<1$, but I get stuck now completing the proof.

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Have you studied infinite series already? I ask because of your tagging... –  DonAntonio Mar 17 '13 at 6:08
    
@DonAntonio I have prior to this class, but the book asks to prove it by definition by using neighborhoods and such. Maybe I should add the real analysis tag as well. –  Q.matin Mar 17 '13 at 6:09
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I think the tagging is fine, and I asked about series because with them the proof of your question is almost immediate. –  DonAntonio Mar 17 '13 at 6:10
    
For all sufficiently large $n$, say $n > N$, we get $x_{n+1}/x_n < r$. So $x_{n+1} < x_n r$. By induction we get $x_{n+1} < x_nr < x_{n-1}r^2 < \cdots < x_N r^{n-N}$. Now the claim follows with $C = x_N/r^N$. –  William Mar 17 '13 at 6:19
    
@William Will induction be the only way to show that $C = x_N/r^N$? Is there a way to show it by using definitions of neighborhoods, bounded? –  Q.matin Mar 17 '13 at 6:22
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2 Answers

up vote 4 down vote accepted

You write:

Since I know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$ and that $0<r<1$, then I have $0<\lim\left(\frac{x_{n+1}}{x_n}\right)\le r<1$, but I get stuck now completing the proof.

You do know that $\lim\left(\frac{x_{n+1}}{x_n}\right)<1$, but you do not know that $0<r<1$: you are supposed to be proving that there is some real number $r$ such that $0<r<1$ (and $r$ has other properties as well).

Start by using the definition of the limit of a sequence. Let $L=\lim_{n\to\infty}\frac{x_{n+1}}{x_n}$. Then by the definition of limit you know that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $$\left|\frac{x_{n+1}}{x_n}-L\right|<\epsilon$$ for all $n\ge n_\epsilon$. Let $r=\frac12(L+1)$; then $L<r<1$. (Why?) Let $\epsilon=r-L$; certainly $\epsilon>0$, so we know that $$\left|\frac{x_{n+1}}{x_n}-L\right|<\epsilon\tag{1}$$ for all $n\ge n_\epsilon$. But the inequality $(1)$ is equivalent to

$$L-\epsilon<\frac{x_{n+1}}{x_n}<L+\epsilon=r\;,$$

so we now know that $\dfrac{x_{n+1}}{x_n}<r$ for all $n\ge n_\epsilon$. Let

$$C=\frac{x_{n_\epsilon}}{r^{n_\epsilon}}\;;$$

Then $x_{n_\epsilon}=Cr^{n_\epsilon}$, and I leave it to you to show that $x_n<Cr^n$ for all $n>n_\epsilon$. Since $0<r<1$, we know that $\lim_{n\to\infty}Cr^n=0$, and since $0<x_n<Cr^n$ for all $n>n_\epsilon$, it then follows from the squeeze theorem that $\lim_{n\to\infty}x_n=0$.

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Beautiful! Proceeding with the correct logical terminology is what I was having the most trouble with, but this is going to be very helpful in providing me a starting path to proceed with these types of questions. Thanks a lot! –  Q.matin Mar 17 '13 at 6:35
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@Q.matin: You’re welcome! –  Brian M. Scott Mar 17 '13 at 6:36
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@Q.matin: Any number strictly between $L$ and $1$ would work equally well; I just chose the number halfway between because it was easy. (Remember, the arithmetic mean of two numbers is halfway between them.) –  Brian M. Scott Mar 17 '13 at 7:08
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@Q.matin: I defined $C$ by the formula $$C=\frac{x_{n_\epsilon}}{r^{n_\epsilon}}\;;$$ just multiply both sides by the denominator. –  Brian M. Scott Mar 22 '13 at 5:37
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@Q.matin: Excellent! You’re welcome. –  Brian M. Scott Mar 22 '13 at 5:53
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Write the definition of $\ell=lim_{n\to+\infty}\frac{x_{n+1}}{x_n}$ and choose $\varepsilon$ such that $r=\ell+\varepsilon<1$.

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