I need to show that for $0 < q < 1$ that the sequence ${x_n} = q^n$ converges, and to find its limit.

So I need to show that

$\forall \epsilon > 0$ there is $N \in \mathbb{N}$ such that $\forall n>N, |x_n - x_m | < \epsilon$ , or $|q^n - q^m| < \epsilon \\$

Once I show convergence it should be pretty easy to find its limit. Don't know where to start with this though.


Hint: You are trying to prove that the sequence is a Cauchy sequence. It is easier to prove directly from the definition of limit that $\lim_{n\to\infty} q^n=0$.

We have $q=\frac{1}{1+p}$ for some positive $p$. Now by the Binomial Theorem, we have $(1+p)^n \gt np$ if $n\ge 1$. This inequality is enough for a proof based on the $\epsilon$-$N$ definition of limit.

The inequality is also enough to show that the sequence $(q^n)$ is Cauchy, if you want to take that path.

Added: For the inequality $(1+p)^n \ge np$, we can use the Binomial Theorem $$(1+p)^n =1+\binom{n}{1}p+\binom{n}{2}p^2+\cdots +p^n.$$ In particular, $(1+p)^n\ge np$, since $np$ is one of the terms in the above expansion.

Now suppose we are given $\epsilon\gt 0$. We have $$|q^n-0|=q^n =\frac{1}{(1+p)^n}\le\frac{1}{pn}.$$ To make $q^n$ less than $\epsilon$, it is enough to make $\frac{1}{np}\lt \epsilon$, that is, make $n\gt \frac{1}{p\epsilon}$. Thus if $N$ is say the smallest integer greater than $1/(p\epsilon)$, and $n\ge N$, then we will have $q^n\lt \epsilon$.

  • $\begingroup$ yeah, the thing is, the way the problem is set up seems it wants me to prove it converges first BEFORE exhibiting the limit. what do you think? $\endgroup$ – terrible at math Feb 19 '14 at 0:38
  • $\begingroup$ also, can you explain why we have $q = \frac{1}{1+p} $? $\endgroup$ – terrible at math Feb 19 '14 at 0:44
  • $\begingroup$ We want to find $p$ so that $q=\frac{1}{1+p}$, or equivalently so that $q(1+p)=1$, or equivalently so that $qp=1-q$. So we take $p=\frac{1-q}{q}=\frac{1}{q}-1$. This approach is one we can take before the logarithm has been introduced and the basic properties of exponential, logarithm proved. After we officially have log, it is more direct to say we want $q^n\lt \epsilon$, so we want $n\gt \frac{\ln \epsilon}{\ln q}$. $\endgroup$ – André Nicolas Feb 19 '14 at 2:02
  • $\begingroup$ Or maybe you were asking why use this trick? Because it works nicely. It is easier to examine how $(1+p)^n$ behaves for large $n$ than to examine directly how $q^n$ behaves. $\endgroup$ – André Nicolas Feb 19 '14 at 2:04
  • $\begingroup$ Thanks a lot! very helpful as usual. $\endgroup$ – terrible at math Feb 19 '14 at 2:11

You could think backwards and "guess" the limit. If $q$ is say, 0.2, what is $0.2^{1000}$? It is approximately $0$, so let us use this guess.

We want to show that there is a threshold $N$ beyond which all $n$ satisfy: \begin{align} |q^n-0|<\epsilon\\ q^n<\epsilon \end{align}

So, if you have proved existence and properties of the logarithm, if you let $N=\log_q\epsilon $, for all $n>N$,

$$q^n<q^N=q^{\log_q \epsilon}=\epsilon$$

Which is exactly what we wanted.

If you haven't proved existence of the logarithm, you can use variants of the Archimedean Property to arrive at the same result.

  • $\begingroup$ just out of curiosity, how would I use the archimedean property? $\endgroup$ – terrible at math Feb 19 '14 at 2:23

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