# Define the sequence of rational numbers p(n) recursively as follows:

$p(1)=1$

$p(n)=1-\frac{p(n-1)}{2}$

• Compute and plot p(n) for $$n=1,2,\ldots , 20$$. What inference do you draw about the terms p(n)?

Replace this text with your caption

# Define the sequence of rational numbers a(n) recursively as follows:

$a(1)=1$

$a(n)=\frac{1+1}{(1+a(n-1))}$

• Compute a(n) and $$a(n)^2-2$$ for $$n=1,2,\ldots , 20$$. What inference do you draw about the terms $$a(n)^2-2$$?

The graph is different with points that are increasing and decreasing. $$a(n)=\frac{1+1}{(1+a(n-1))}$$ is a horizontal line while $$a(n)^2-2$$ is a more complex solution with a different inf(a(n)) and sup(a(n)).

• From the recursive definition of p(n) find an equation for, and so evaluate, a possible limit p for p(n) as n increases without bound. $$p(1)=1$$

$$p(n)=1-\frac{p(n-1)}{2}$$

$\lim_{n\to\infty} p(n) = p$

exists then p satisfies $p = 1 - \frac{p}{2}$ $p+\frac{p}{2} = 1$ $\frac{3p}{2} = 1$ $\frac{2}{3}*\frac{3p}{2} = 1*\frac{2}{3}$ $p = \frac{2}{3}$

• Show, from the recursive definition of p(n) that $$\vert p(n) - p\vert = \frac{1}{2}\vert p(n-1) - p\vert$$ for all $$n \geq 1$$. $p(n)=1-\frac{p(n-1)}{2}$

$p(n)-\frac{2}{3} = 1 - \frac{p(n-1)}{2} - \frac{2}{3}$

$\frac{1}{3} - \frac{p(n-1)}{2} = -\frac{1}{2}(p(n-1)-\frac{2}{3})$

Hence $\vert p(n) - p \vert = \frac{1}{2} \vert p(n-1) - p \vert$

• So show that $$\vert p(n) - p\vert \leq (\frac{1}{2})^{n-1}$$ for all $$n \geq$$ 1.

We have $p(1) - p = \frac{1}{3}$ so its true for n = 1

Assume that, for some $$n \geq 1$$ $$\vert p(n) -p \vert \leq (\frac{1}{2})^{n-1}$$ is true

Then $\vert p(n+1) - p \vert = \frac{1}{2}\vert p(n-1) - p \vert \leq \frac{1}{2}^{n}$

Then we have $\vert p(n) - p \vert \leq \frac{1}{2}^{n-1}$ for all $$n \geq 1$$.

• Prove that $$2^{n-1} \geq n$$ for all $$n\geq 1$$.

This is true fr n = 1.
Assume it’s true for some n.

Then $$2^n = 2 * 2^{n-1} \geq 2n \geq n+1$$ since $$n \geq 1.$$
Therefore, by induction, $$2^{n-1} \geq n$$ for all $$n \geq 1$$

• Prove that the sequence p(n) converges to p.

Notice that for all $$n \geq 1$$, $\vert p(n) - p \vert \leq \frac{1}{3}(\frac{1}{2})^{n-1} \leq \frac{1}{3n}$

so given $$\varepsilon > 0$$ if we choose $$N > \frac{1}{3\varepsilon}$$ then $$\vert p(n) -p\vert \leq$$ all $$n \geq N$$

which means that

$\lim_{n\to\infty} p(n) = p= \frac{2}{3}$