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Definition: A sequence $(a_n)$ converges to a real number $a$ if, for every positive $\epsilon$, there exists an $N \in \mathbb N$ such that whenever $n \ge N$ it follows that $|a_n - a| \lt \epsilon$.

Using only this definition prove that if $(x_n) \rightarrow 2$ then $$(\frac{2x_n -1}{3}) \rightarrow1.$$

Solution. Since $(x_n) \rightarrow 2$ then there exists $N_1 \le n$ such that $|x_n -2| \lt \epsilon$. Choose $N_2 \le n$ such that $|x_n -2| \lt \frac{3\epsilon}{2}$. Then we have $$\frac{2}{3}|x_n -2| \lt \epsilon \Rightarrow \frac{2x_n -4}{3} \lt \epsilon $$

from which follows $$\frac{2x_n -1 -3}{3} \lt \epsilon $$
and hence $$\left|\frac{2x_n -1}{3} -1\right| \lt \epsilon $$ which by definition is $$\left(\frac{2x_n -1}{3}\right) \rightarrow1.$$

Is my argument correct to prove the fact? I feel like its missing something. Thank you in advance.

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    $\begingroup$ what is $n$??.. $\endgroup$ – mathworker21 Jul 14 '18 at 5:18
  • $\begingroup$ @mathworker21 Thats the thing... How do I write $n$ less than some $\epsilon$ value in this case? $\endgroup$ – Red Jul 14 '18 at 5:22
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So you are given an assumption of the form $\forall x_1 \exists y_1 ~P$, and you need to prove a statement of the form $\forall x_2 \exists y_2 ~ Q$. To prove the conclusion, you must present a function $y_2(x_2)$ that satisfies $Q$, and you may assume the existence of the function $y_1(x_1)$ which satisfies $P$. For this problem, you have to do that by defining $y_2$ in terms of $y_1$.

Since $(x_n) \to 2$ then there exists $N_1 \le n$

No. $N$ is a function of $\epsilon$, of the form $\mathbb R \to \mathbb N$. $N$ is not a function of $n$. If you want to call it $N_1$ to distinguish it from the solution that is good though.

Choose $N_2 \le n$ such that $|x_n−2| < \frac{3\varepsilon}2$.

That doesn't make sense to me. You are correct that the core of the proof is to show that $N_2 : \mathbb R \to \mathbb N$ exists which satisfies the conclusion. But I don't see how you are defining that here. You need to say something like

"Choose $N_2$ such that $N_2(\varepsilon) = \dots$"

And for this problem it is that $N_2(\varepsilon)= N_1(\frac 32 \varepsilon)$. So basically you go:

To prove: $$n > N_2(\varepsilon) \implies \left|\frac{2x_n - 1}3 - 1\right| < \varepsilon$$

Choose $N_1(\frac 32 \varepsilon)$ as the witness for $N_2(\varepsilon)$ so: $$n > N_1\left(\frac 32 \varepsilon\right) \implies \left|\frac{2x_n - 1}3 - 1\right| < \varepsilon$$

You can change variables and let $3/2~\varepsilon = d$:

$$n > N_1(d) \implies \left|\frac{2x_n - 1}3 - 1\right| < \frac 23 d$$

Some real number equivalences:

$$n > N_1(d) \implies \left|x_n - 2\right| < d$$

And that is the original assumption, which shows that choosing $N_1(\frac 32 \varepsilon)$ as the definition for $N_2(\varepsilon)$ satisfies the conclusion.

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You've shown that, for sufficiently large $n$, \begin{align*} \frac{2x_n-1}{3} < 1+\varepsilon. \end{align*} You would also need to show that for large enough $n$, \begin{align*} \frac{2x_n-1}{3} > 1-\varepsilon. \end{align*} However, the proof is trivial if we simply invoke the continuity of the function $f(x) = (2x-1)/3$.

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