Exercise 6.3.3 (Monotone bounded sequences converge)

Question

Prove Proposition 6.3.8.\par 

	extbf{Proposition 6.3.8.} Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers which has some finite upper bound $M \in \mathbf R$, and which is also increasing (i.e., $a_{n+1} \geq a_n for all n \geq m$). Then $(a_n)_{n=m}^\infty$ is convergent, and in fact

$$\displaystyle \lim_{n	o\infty} a_n = \sup(a_n)_{n=m}^\infty \leq M.$$

Prakash Anand · Accepted Answer

\begin{proof}
$\left\{a_{n}\right\}$ is a sequence, increasing in $n$, that is bounded above by $M$ finite. By the definition of $L=\sup \left\{a_{n}\right\} L \leq M$. In particular, $L$ is finite. Fix $\epsilon \gneq 0$ and consider $L-\epsilon$. By definition of $L$ this is no longer an upper bound, so there exists $N$ such that $L-\epsilon \leq a_{N} \leq L$. But $\left\{a_{n}\right\}$ is increasing and $L$ is the supremum imply $L-\epsilon \leq a_{n} \leq L$, $\forall n \geq N$. $\epsilon$ was chosen arbitrarily, so our sequence is eventually $\epsilon$ close to its sup for any $\epsilon \gneq 0$.
\end{proof}

Exercise 6.3.3 (Monotone bounded sequences converge)

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Exercise 6.3.3 (Monotone bounded sequences converge)

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