Proof that $0.33333... = \frac{1}{3}$ using $\epsilon-N$ method

Question

This proof is quite prevalent on the web, yet I struggle using this particular method.

Wikipedia (http://en.wikipedia.org/wiki/Limit_of_a_sequence) tells us:

We call $x$ the limit of the sequence $\{ x_n\}$ if:

for each real number $\epsilon> 0$ , there exists a natural number $N$ such that, for every natural number $n > N$, we have $|x_n - x| <\epsilon$.

So I start with the sequence $\{ x_n\} = \{0.3, 0.33, 0.333, 0.3333, \ldots\}$.

Then I want $\left|x_n - \dfrac{1}{3} \right| < \epsilon$, since this would mean that $\dfrac{1}{3}$ is the limit of the sequence $\{x_n\}$, by definition:

$$\left|x_n - \frac{1}{3} \right| = \left|\frac{1}{3} - x_n\right| = \dfrac{1}{3} - x_n = \frac{1}{3}10^{-n}$$

So we want $\dfrac{1}{3}10^{-n}<\epsilon$.

Where do I go from here? Where does the $N$ come in?

And also: can I simply say that $0.33333\ldots$ is the limit of the sequence $\{x_n\}$. By theorem a sequence can have at most one limit, thus this must mean that $\dfrac{1}{3} = 0.33333\ldots$, since both are limits of the sequence.

Answer 1

Find $N> -\log_{10}(3\epsilon)$. Then if $n>N$, then $10^n>\frac{1}{3\epsilon}$ or $\frac{1}{3}\cdot 10^{-n}<\epsilon$.

Answer 2

The $N$ comes in as an "interval around infinity." The definition says that there is an $N$ such that for all $n> N$ (i.e., for all $n\in(N,\infty)$), then the final inequality holds.

What you have done so far is say that it must be the case that $\frac{1}{3}10^{-n}<\varepsilon$, which is the same as saying $n>-\log_{10}(3\varepsilon)$. If you let $N$ be a positive integer such that $N>-\log_{10}(3\varepsilon)$, then it follows that whenever $n>N$ then $\frac{1}{3}10^{-n}<\epsilon$.

Answer 3

You did no justify your value for $x_n$.

The value of a decimal (base 10) string with infinite number of digits 3 behind the period is: $$ (0.333\cdots)_{10} = \sum_{k=1}^{\infty} 3\cdot 10^{-k} = 3 \sum_{k=1}^{\infty} 10^{-k} = 3 \sum_{k=1}^{\infty} \left(\frac{1}{10}\right)^k = 3 \lim_{n\to \infty} S_n $$ for the partial sums $$ S_n = \sum_{k=1}^{n} \left(\frac{1}{10}\right)^k $$ In your notation $x_n = 3 S_n$. Their value can be evaluated by the geometric sum in $q = 1/10$. $$ S_n = \frac{1 - (1/10)^{n+1}}{1 - (1/10)} - 1 \\ = \frac{(10^{n+1} - 1)/10^{n+1}}{9/10} - 1 \\ = \frac{10^{n+1} - 1}{9\cdot 10^n} - \frac{9\cdot 10^n}{9\cdot 10^n} \\ = \frac{10^n - 1}{9\cdot 10^n} \\ = 1/9 - 1/(9 \cdot 10^n) $$ This means $x_n = 1/3 - 1/(3\cdot 10^n)$ and $1/3 - x_n = 1/(3\cdot 10^n)$, which agrees with your value.

For any challenge $\epsilon > 0$ we have $$ 1/(3\cdot 10^n) < \epsilon \Rightarrow \\ 1/(3 \cdot \epsilon) < 10^n \Rightarrow \\ \ln(1/3 \cdot \epsilon) < n \ln(10) \Rightarrow \\ N := \frac{\ln(1/3 \cdot \epsilon)}{\ln(10)} < n $$

Answer 4

Define $x_n=\dfrac{10^n-1}{3.10^n}$. Then establish $\left|x_n-\dfrac13\right|=\left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|$.

And so, $\left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|=\left|\dfrac{10^n-1-10^n}{3.10^n}\right|=\left|\dfrac{-1}{3.10^n}\right|=\left|\dfrac{1}{3.10^n}\right|\lt\epsilon\;\forall n\ge N(\epsilon)$