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In my homework set, I have the following question:

Show that $$ a_n = \frac{\sin(a_{n-1}) + 1}{2}, \quad a_1=0 $$

satisfies the definition of Cauchy sequence.

As we went over the concept of Cauchy sequences a bit too quickly in class, I'm puzzled about how I should go about showing this. Conceptually, I understand what a Cauchy sequence is, and I know that in $\mathbb{R}$ it is the same as convergence, but I find it hard to apply to the above sequence.

I'd appreciate some hints about how to approach this problem.

Edit: I've been working on this with help from responses below so I thought I'd update my work. It's verbose but I thought that might show my thinking process better.

My work so far:

$$ \begin{align} |a_{n+1} - a_n| &= \left | \frac{\sin(a_n) + 1}{2} - \frac{\sin(a_{n-1}) + 1}{2} \right | \\ \\ &= \frac{1}{2} \left | \sin(a_n) + 1 - (\sin(a_{n-1}) + 1)\right | \\ \\ &= \frac{1}{2} \left | \sin(a_n) - \sin(a_{n-1}) \right | \\ \\ &= \frac{1}{2} \left | 2 \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\ \\ &\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \times \cos(\frac{a_n + a_{n-1}}{2}) \right | \\ \\ &\leq \frac{1}{2} \left | \sin(\frac{a_n - a_{n-1}}{2}) \right | \\ \\ &\leq \frac{\left | a_n - a_{n-1} \right |}{2} \end{align} $$

This set of inequalities work because $|sin(x)| \leq |x|, \ \forall x \in \mathbb{R}$.

Hence, we have:

$$ \begin{align} |a_{n+1} - a_n| &\leq \frac{1}{2} \left |a_n - a_{n-1} \right | \leq \frac{1}{2^2} \left | a_{n-1} - a_{n-2} \right | \leq \dots \leq \frac{1}{2^{n-1}} \left |a_{2} - a_{1} \right | \end{align} $$

So we learn that the distance between consecutive terms is becoming smaller, and that:

$$ \begin{align} \left | a_{n+1} - a_n \right | &\leq \frac{1}{2^{n-1}} \times \left | \frac{1}{2} - 0 \right | = \frac{1}{2} \times \frac{1}{2^{n-1}} = \frac{1}{2^n} \end{align} $$

Now, for $n>m$ and $k=m$ we have:

$$ \begin{align} \left | a_n - a_m \right | &= \sum_{k=m}^{n-1} \left | a_{k+1} - a_k \right | \\ \\ &= \sum_{k=m}^{n-1} \frac{1}{2^k} \\ \\ &= \left ( \frac{1}{2^m} + \frac{1}{2^{m+1}} + \dots + \frac{1}{2^{n-2}} + \frac{1}{2^{n-1}} \right ) \\ \\ &= \frac{1}{2^{m-1}} \left ( \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-m-2}} + \frac{1}{2^{n-m-1}} \right ) \\ \\ &\leq \frac{1}{2^{m-1}} \end{align} $$

So, if we have $|a_n - a_{n-1}| < \epsilon$, then $\frac{1}{2^{m-1}} < \epsilon$. If we solve for $m$, we get that $m > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$ and therefore $N > \frac{\log(\frac{2}{\epsilon})}{\log(2)}$.

So we have found $|a_n - a_m| < \epsilon$ for $n, m > N$. Hence, $(a_n)$ is Cauchy.

Resources used:

  1. Answers given below.
  2. This video was very helpful
  3. Understanding the definition of Cauchy sequence
  4. Proving that a sequence such that $|a_{n+1} - a_n| \le 2^{-n}$ is Cauchy
  5. Showing a recursive sequence is Cauchy
  6. How do I find the limit of the sequence $a_n=\frac{n\cos(n)}{n^2+1}$ and prove it is a Cauchy sequence?
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  • $\begingroup$ @RobertZ: I've updated my attempt. What do you think of this? $\endgroup$ – JHG90 Oct 22 '17 at 15:06
  • $\begingroup$ OK. I updated my answer. Your proof is correct! $\endgroup$ – Robert Z Oct 22 '17 at 15:18
  • $\begingroup$ Awesome! Thanks for your help! $\endgroup$ – JHG90 Oct 22 '17 at 15:33
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Hint. Note that the function $f:\mathbb{R}\to \mathbb{R}$, $$f(x)=\frac{\sin(x)+1}{2}$$ is a contraction: for any $x,y\in \mathbb{R}$ there exists $t$ between $x$ and $y$ such that $$|f(x)-f(y)|\leq \left|\frac{\cos(t)}{2}\right||x-y|\leq \frac{|x-y|}{2}\tag{1}$$ where we used the Mean value theorem. Hence, for $n\geq 1$, by using (1) $(n-1)$ times, we obtain $$|a_{n+1}-a_n|\leq \frac{1}{2}|a_{n}-a_{n-1}|\leq \dots\leq \frac{1}{2^{n-1}}|a_{2}-a_{1}|.$$ Can you take it from here?

P.S. If the Mean value theorem is not allowed then we may note that by the sum-to-product identity, $$\frac{\sin(x)+1}{2}-\frac{\sin(y)+1}{2}=\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right).$$ Since $|\sin(t)|\leq |t|$ and $|\cos(t)|\leq 1$, it follows that (1) holds.

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    $\begingroup$ Note that $a_{n+1}=f(a_n)$ then by the Contraction Theorem the sequence $(a_n)_n$ converges to the unique fixed point which is $1/2$. Take a look at the proof of the CT en.wikipedia.org/wiki/Banach_fixed-point_theorem . They show that the sequence is a Cauchy sequence. $\endgroup$ – Robert Z Oct 20 '17 at 13:49
  • $\begingroup$ One more question..Why you take the particular interval $[-1/2,1/2]$ ? $\endgroup$ – Empty Oct 20 '17 at 13:54
  • $\begingroup$ We need an invariant set, but in this case, you are right, no need of restariction.We can take the whole real line. $\endgroup$ – Robert Z Oct 20 '17 at 13:58
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    $\begingroup$ Yes, both closed sets $[-1/2,1/2]$ and $\mathbb{R}$ are complete. However here OP need only to show the Cauchy sequence property so an invariant set is sufficient. $\endgroup$ – Robert Z Oct 20 '17 at 14:02
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    $\begingroup$ MVT is not strictly necessary here, just rewrite the difference $\frac{\sin(x + 1)}{2}-\frac{\sin(y + 1)}{2}$. See my P.S. $\endgroup$ – Robert Z Oct 21 '17 at 8:44
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I think it is not necessary to define such a function defined by Robert Z.

$\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}|a_n-a_{n-1}|$ for all $n=1,2,\cdots$

So , $\displaystyle |a_{n+1}-a_n|\le \frac{1}{2}.|a_n-a_{n-1}|\le \frac{1}{2^2}.|a_{n-1}-a_{n-2}|\le \cdots \le \frac{1}{2^{n-1}}.|a_2-a_1| \text{ for all } n=1,2,\cdots$

Now , for any positive integer $p$ , $$|a_{n+p}-a_n|\\\le |a_{n+p}-a_{n+p-1}|+\cdots +|a_{n+2}-a_{n+1}|+|a_{n+1}-a_n|\\\le \left( \frac{1}{2^{n+p-2}}+\cdots+\frac{1}{2^{n}}+\frac{1}{2^{n-1}}\right).|a_2-a_1|\\=\frac{1}{2^{n-2}}.[1-(1/2)^p].|a_2-a_1|\\ \to 0 \text { , as } n\to \infty \text{ for all } p=1,2,\cdots $$

So , $\{a_n\}$ is Cauchy.

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