# Monotone bounded sequences; Cauchy sequences

#### Question 1(a)

Show that the sequence x(n) defined recursively as follows:

$x(1)=1$ $x(n)=x(n-1)+ \frac{1}{n^2}$

is (monotone) increasing and bounded above by 2 and therefore converges.

Sequence x(n) or real numbers is monotone increasing if $$x(n+1)\geq x(n)$$ for all $$n\geq 1$$.

Since $$\frac{1}{n^2}$$ always equals a positive integer and as $$\lim_{n\to\infty} \frac{1}{n^2}$$ is increasing for $$n \geq 1$$ we have if $$x(n+1)\geq x(n)$$.
$$\therefore x(n)$$ is monotone increasing.
The sequence $$x(n)$$ is bounded above if there is a real number U such that $$x(n)\leq U$$ for all $$n\geq 1$$.

Lets say $$U=3$$ then $$x(n) \leq 3$$ for all $$n\geq 1$$.
Since $$\frac{1}{n^2}$$ is a fraction that is decreasing as n $$\rightarrow \infty$$ and x(1) = 1 we have a convergence and x(n) is bounded above by 2.

#### (b)

Calculate and plot x(n) for $$1\leq n \leq 1000$$ and so get a rough approximation to $$\lim_{n\to \infty}x(n)$$.

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#### Question 3(a)

Is the sequence x(n) defined recursively as follows:

$x(1)=1, x(2)=2;$ $x(n)=\frac{x(n-1)+x(n-2)}{2}$
monotone increasing?
monotone decreasing?
bounded above?
bounded below?

A sequence x(n) or real numbers is monotone increasing if $$x(n+1)\geq x(n)$$ for all $$n\geq 1$$.

At first we stated $$x(2) = 2 > x(1) =1$$ so the sequence starts monotone increasing.
When x = 3 we get $x(3) = \frac{x(3-1) +x(3-2)}{2}$ $= \frac{x(2)+x(1)}{2}$ $=\frac{2+1}{2}$ =$$\frac{3}{2}$$

Notice that $$x(2) > x(3) > x(1)$$.

So x(n) is monotone increasing and decreasing for all $$n\geq 1$$.

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