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I'm reading Courant's What is Mathematics?

In page $64$, he gives the example of limit of the successive powers of $q$. If $-1<q<1$ then the successive powers of $q$ will approach zero as $n$ increases. Then he suggests that to give a rigorous proof of that assertion, one needs to use the inequality proved on page $15$, which is:

$$(1+p)^n\geq 1+np$$

But I'm not sure on how these things are connected - And I guess I'm unable to provide more information on my doubt, I just want to understand why that inequality is used to prove that that limit approaches $0$ as $n\rightarrow \infty$ for $-1<q<1$.

I have something to add to this question. I guess I'm half the way to understand , but this is not the point. Reading the sentence again:

If $-1<q<1$ then the successive powers of $q$ will approach zero as $n$ increases.

I have an argument that could work as a proof, and it's a lot simpler: Suppose we take the number $10$, observation the successive powers of 10:

$$ \begin{array}{cr} 10 & 10 \\ 10^2 & 100 \\ 10^3 & 1000 \\ 10^4 & 10000 \\ \end{array}$$

As we raise $n$ in $10^n$, we add one zero and the one goes on sliping to the left, this will happen for any number bigger than $1$ and different of $0$, the process will only take a little more time to happen, for $2^n$ we have:

$$ \begin{array}{cc} 2 & 2 \\ 2^2 & 4 \\ 2^3 & 8 \\ 2^4 & 16 \\ \end{array}$$

If $a$ is smaller than $1$ in $a^n$, then we'll have for example:

$$\begin{array}{cc} \frac{1}{10} & 0.1 \\ \left(\frac{1}{10}\right)^2 & 0.01 \\ \left(\frac{1}{10}\right)^3 & 0.001 \\ \left(\frac{1}{10}\right)^4 & 0.0001 \\ \end{array}$$

Then if the numbers are different of $1$ and $0$, I know that these numbers are going to walk. If $n<1$, it's going to walk to the right, and if $n>1$, the number will walk to the left. This looks pretty convincing to me so, why do I need to use Bernoulli's inequality? The answers given by John and RobJohn are useful, they tell me how to do it. But it's not really clear why I should use it instead of the method I just pointed.

Red Banana
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    When you say numbers are "slipping to the left" or "walking", it seems that you are saying $$a_{n+1}\lt a_n$$ However, this is true of the sequence $$a_n=\frac12+\frac1n$$ but that sequence does not tend to $0$. – robjohn Jan 17 '14 at 15:41

2 Answers2

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Let $|q| <1$, $q\neq 0$. Let $\frac{1}{|q|} =1+ b$ for some positive $b$. Then

$\frac{1}{|q|^n} = (1+b)^n \geq 1 + bn$. Which implies

$|q^n| \leq \frac{1}{1+nb}$. As we know the right hand side goes to zero, so is the left hand side.

  • Sorry. Why $\frac{1}{|q|} =1+ b$? I guess that it's one equal to the other in order to adapt to the inequality and hence use it to prove what's desired. – Red Banana Oct 16 '13 at 05:35
  • As $|q|<1$, $\frac{1}{|q|}>1$. Then $b= \frac{1}{|q|}-1 >0$. –  Oct 16 '13 at 05:36
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One way to show that $q^n\to0$ when $|q|\lt1$ is to look at logarithms. That is, $\log|q|\lt0$, so $$ \begin{align} \log\left(\lim_{n\to\infty}|q|^n\right) &=\lim_{n\to\infty}n\log(q)\\ &=-\infty \end{align} $$ which essentially says that $$ \lim_{n\to\infty}|q|^n=0 $$

However, Bernoulli's Inequality can be proven without using logarithms, and we can use that instead.

Suppose that $0\lt q\lt1$, then $\frac1q\gt1\iff r=\frac1q-1\gt0$. Then Bernoulli says that $$ \begin{align} q^n &=\frac1{(1+r)^n}\\ &\le\frac1{1+nr} \end{align} $$ Since $r\gt0$, the Archimedean Property of the reals says that there is a $k$ so that $kr\gt1$. Then $$ \begin{align} q^{nk} &\le\frac1{1+nkr}\\ &\le\frac1n \end{align} $$ ans therefore, we can deduce that $$ \lim_{n\to\infty}q^n=0 $$ This maintains the spirit of the logarithmic proof without using logarithms.

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robjohn
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  • Would the downvoter care to comment? – robjohn Oct 16 '13 at 17:48
  • I downvoted because your answers looks essentially the same as the other one. – Hu Zhengtang Oct 16 '13 at 17:53
  • @Giraffe: thanks for explaining. I really appreciate it. Since there is not much latitude given Bernoulli's Inequality and what we are trying to prove, the equations would look similar. However, I thought that I was adding some useful context and explanation that was missing in the other answer. – robjohn Oct 16 '13 at 18:02
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    You are welcome. I agree that your answer contains much more details, which is quite helpful for beginners; however, personally I wouldn't like to post an answer if it has no essential difference from existing answers. – Hu Zhengtang Oct 16 '13 at 18:26
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    @Giraffe: I disagree that there is no essential difference. The question was not how to apply Bernoulli (which I agree, both John and I showed), but why to apply Bernoulli. I believe that the notion that for $|x|\lt1$ we have $x^n\to0$ is used almost axiomatically, and the reason for the question was that the OP was wondering why use Bernoulli for something so obvious. Thanks for your comment. – robjohn Oct 16 '13 at 19:54