Monotone bounded sequences; Cauchy sequences

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.

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

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) +

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