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If pn is a non-decreasing sequence and there is a point, p, to the right of each point of the sequence, then the sequence converges to some point.
That is an oddly stated problem.Originally Posted by spikedpunch
Mathematically translated it says “A non-decreasing sequence bound above converges”.
Which of course is true. The proof follows from the completeness property.
The sequence converges to its least upper bound.
Isn't this the Monotone convergence theorem?
"If a sequence is monotone and bouned, then is converges"
here is an outline:
Suppose the sequence $\displaystyle (p_{n})$ is increasing and bounded above.
Consider the set of all points in the sequece $\displaystyle \{p_{n}|n\in\mathbb{N}\}$.
By the axiom of completeness, $\displaystyle sup\{p_{n}|n\in\mathbb{N}\}$ exists (since the set is a bounded subset of $\displaystyle \mathbb{R}$.
Assume $\displaystyle lim(p_{n})=s$, where $\displaystyle s=sup\{p_{n}|n\in\mathbb{N}\}$.
Then, for $\displaystyle \epsilon>0$, $\displaystyle s-\epsilon<s$. Thus, there exists a point $\displaystyle p_{N}$ such that $\displaystyle s-\epsilon<p_{N}<s$. Furthermore, because our sequence is increasing, it follows that, for every $\displaystyle n\geq\\N$, $\displaystyle p_{N}\leq\\p_{n}$. Thus, $\displaystyle s-\epsilon<p_{N}\leq\\p_{n}<s<s+\epsilon$.
This implies $\displaystyle |p_{n}-s|<\epsilon$
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