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Starting from $0<a_1\le b_1$, replace the lower endpoint by the geometric mean and the upper endpoint by the arithmetic mean. The arithmetic-geometric mean inequality gives $a_n\le a_{n+1}\le b_{n+1}\le b_n$. The bounded monotone sequence theorem gives limits, and the arithmetic recurrence forces them to coincide. The limiting value is the arithmetic-geometric mean of the initial pair.

Even restricting to increasing sequences in $[0,1]$ with limit of a sequence one gives no common stopping index. For any positive integer $K$, defineThis is an increasing bounded sequence converging to one, but its absolute error is one through index $K$. Since $K$ can exceed any proposed index, the bounded monotone sequence theorem alone cannot say when the absolute error will fall below $1/2$. A zero difference between two successive terms is also insufficient: the initial plateau eventually ends.

There is no universal eventual rate either. Let $(r_n)$ be any proposed positive nonincreasing sequence tending to zero, with $r_n\le 1$. SetThen $(a_n)$ is increasing, lies in $[0,1]$, and converges to one, but for every sufficiently large $n$ its absolute error satisfies $1-a_n=\sqrt{r_n}>r_n$. Indeed, $\sqrt{r_n}/r_n=1/\sqrt{r_n}\to \infty$, so no fixed constant multiple of $r_n$ bounds the absolute error eventually. If a proposed rate is not nonincreasing, replace it, after a finite initial segment if necessary, by the decreasing envelope $s_n=\sup \{r_k:k\ge n\}$; this also tends to zero, and the same construction defeats the original rate. This is an original construction, rather than a quantitative claim attributed to the qualitative theorem.

For a concrete slow example, $a_n=1-1/\log (n+2)$ increases to one, where $\log$ is the natural logarithm. Its absolute error is $1/\log (n+2)$, so making it less than $\epsilon$ requires $n+2>\exp (1/\epsilon )$, using the exponential function. At $\epsilon =0.1$, this first happens at $n=22025$. The reciprocal and geometric examples in increasing and decreasing approximations to a common limit meet the same tolerance at $n=11$ and $n=4$, respectively. All three obey the same bounded monotone sequence theorem.

If $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. The nondecreasing partial sums of the nonnegative $a_n$ are bounded above by a finite initial contribution plus the sum of $b_n$; bounded monotone sequence theorem proves the claim. Consequently $|c_n|\le b_n$ with summable $b_n$ proves absolute convergence of $\sum c_n$. Finite changes to initial terms do not affect convergence.

For the two sequences in bounded monotone sequence theorem,Their directions and bounds are explicit. One is a succession of lower approximations and the other a succession of upper approximations to the same limit of a sequence. To obtain absolute error strictly below $\epsilon >0$, it suffices to choose an integer $N>1/\epsilon$, and then every $n\ge N$ works. For $\epsilon =0.01$, the first such index is $101$; at $100$ the absolute error equals the tolerance.

A geometric sequence offers a faster example. The recurrence relation $c_1=1/2$, $c_{n+1}=(c_n+1)/2$ gives $c_n=1-2^{-n}$. Thus $(c_n)$ is an increasing bounded sequence with limit of a sequence one and absolute error $2^{-n}$. For $\epsilon =0.01$, $n=7$ already works, since $2^{-7}=1/128<0.01$, whereas $2^{-6}=1/64>0.01$. The bounded monotone sequence theorem guarantees both examples converge; the formulas explain their different speeds.

A real sequence is increasing when $a_{n+1}\ge a_n$ and decreasing when $a_{n+1}\le a_n$.

A monotone sequence has one of these directions for every index. Here increasing means nondecreasing and decreasing means nonincreasing: equal successive terms are allowed. Strict increase would require $a_{n+1}>a_n$ at every step, and strict decrease would require $a_{n+1}<a_n$. A constant sequence is both increasing and decreasing. These conventions are those of Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl (§2.1.1, Definition 2.1.9).

For $n\ge 1$, the sequence $1-1/n$ increases toward one, while $1+1/n$ decreases toward one. Repeating each term gives the still increasing sequence $0,0,1/2,1/2,2/3,2/3,\dots$. Plateaus do not obstruct the bounded monotone sequence theorem. Checking the sign of $a_{n+1}-a_n$ is often the simplest way to verify a direction; finding an upper bound or lower bound is a separate task.

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

For $c>0$ and $a_0>0$, the displayed recurrence relation stays positive and satisfiesAfter the first step, $a_n^2\ge c$, so $a_{n+1}-a_n=(c-a_n^2)/(2a_n)\le 0$. The bounded monotone sequence theorem gives a limit of a sequence $L\ge 0$; the inequality $L^2\ge c>0$ makes $L$ positive, and passage to the limit of a sequence in the recurrence relation gives $L^2=c$. This proves existence of the positive square root without assuming it in the construction. Once $s=\sqrt{c}$ is known, the exact error identity $a_{n+1}-s=(a_n-s{)}^2/(2a_n)$ explains the eventual quadratic convergence.

The intermediate value theorem gives a root $r\in (a,b)$, unique because $f^{\prime }>0$ makes $f$ strictly increasing. We show that Newton root-finding iteration, started at $b$, always stays to the right of $r$. If $x>r$, the mean value theorem gives $f(x)=f^{\prime }(\xi )(x-r)$ for some $\xi \in (r,x)$. Since $f^{\prime \prime }>0$, $f^{\prime }$ is strictly increasing, so $0<f^{\prime }(\xi )<f^{\prime }(x)$. It follows thatInduction gives $r<x_{n+1}<x_n\le b$, so the bounded monotone sequence theorem supplies a limit $\alpha \ge r$. The continuous positive $f^{\prime }$ is bounded away from zero on $[a,b]$, making $F$ continuous. Passing to the limit in $x_{n+1}=F(x_n)$ gives $\alpha =F(\alpha )$ and hence $f(\alpha )=0$. Uniqueness implies $\boxed{\alpha =r}$.

For the rate, apply the Taylor theorem with Lagrange remainder to $f$ between $r$ and $x_n$:With $e_n=x_n-r>0$, rearrangement givesThus the monotone Newton convergence for a strictly convex function has quadratic convergence. More concretely, if $K={\max }_{[a,b]}{f}^{\prime \prime }/[2{\min }_{[a,b]}{f}^{\prime }]$, then $e_{n+1}\le K{e}_n^2$. Once $K{e}_N<1$, induction gives $e_{N+j}\le K^{-1}(K{e}_N{)}^{2^j}$, exhibiting the repeated squaring of small errors.

Rearranging the assumed positive-term ratio inequality givesInduction therefore yields $a_n/b_n\le a_1/b_1$ for every $n\ge 1$. In particularThe positive partial sums of the first series are nondecreasing and bounded above, so the bounded monotone sequence theorem proves that they converge. This is the ratio comparison test, and establishes convergence of $\sum a_n$.

For a nonempty set $S\subseteq \mathbb{R}$, a real number $s$ is its supremum if every $x\in S$ satisfies $x\le s$ and every upper bound $u$ of $S$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $x\in S$ has $s-\epsilon <x\le s$. The least upper bound axiom states that every nonempty set of real numbers bounded above has a real supremum.

Suppose $(a_n)$ is nondecreasing and bounded above. Apply the least upper bound axiom to $S=\{a_n:n\ge 1\}$, obtaining $s=\sup S$. For any $\epsilon >0$, $s-\epsilon$ cannot be an upper bound, so there is $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesso $|a_n-s|<\epsilon$. This is precisely the definition of a convergent sequence. If $(a_n)$ is nonincreasing and bounded below, apply the proved case to $(-a_n)$; its limit is the negative of the infimum of the original terms. Thus the bounded monotone sequence theorem follows directly from the axiom:

The two-variable arithmetic-geometric mean inequality follows from a square:Hence $2\sqrt{ab}\le a+b$, with equality exactly when $a=b$.

For the series, apply that same inequality with $a=a_n$ and $b=n^{-2}$ to obtainThe comparison series converges: its first summand has finite sum by hypothesis, and the inverse-square series converges, for example by the telescoping estimate $n^{-2}\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. Thus the comparison test for series provesEquivalently, its positive partial sums are increasing and bounded above by $\frac{1}{2}\sum a_n+\frac{1}{2}\sum n^{-2}$, so the bounded monotone sequence theorem proves convergence directly. This is the case $p=1$ of weighted square roots of a summable sequence.

Let $F(x)=\sqrt{2+\sqrt{x}}$ for $x\ge 0$. This is an increasing continuous function. Starting at $a_1=\sqrt{2}$, induction gives $\sqrt{2}\le a_n\le 2$: if $a_n\le 2$, then $a_{n+1}\le \sqrt{2+\sqrt{2}}<2$, and every iterate is at least $\sqrt{2}$.

The first step has $a_2>\sqrt{2}=a_1$. Since $F$ is increasing, $a_n\ge a_{n-1}$ implies $a_{n+1}=F(a_n)\ge F(a_{n-1})=a_n$. Thus this sequence is increasing and bounded above. Let $L_{-}={\sup }_n{a}_n$. For every $\epsilon >0$, a term exceeds $L_{-}-\epsilon$; all later terms lie between it and $L_{-}$. This proves convergence, the elementary bounded monotone sequence theorem.

For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt{2}$ because every positive iterate is at least that value. Hence it converges to its infimum $L_{+}$. From $b_2=2$ onward all terms lie in $[\sqrt{2},2]$.

Continuity of $F$ permits passing to each limit in the recurrence, giving $L_{\pm }^2=2+\sqrt{L_{\pm }}$. To prove that these fixed points agree, suppose $r,s\ge \sqrt{2}$ both satisfy that equality. Subtracting the two equations and rationalizing givesThe second factor is positive: $r+s\ge 2\sqrt{2}$, whereas $(\sqrt{r}+\sqrt{s}{)}^{-1}\le 1/(2\sqrt{\sqrt{2}})<2\sqrt{2}$. Thus $r=s$. ThereforeExistence has been supplied by the two convergent sequences, and the subtraction argument gives the required uniqueness in their common interval.

A real sequence $x_n$ converges to $l$ if, for every $\epsilon >0$, there is an integer $N$ such that $|x_n-l|<\epsilon$ for every $n\ge N$. A convergent series ${\sum }_{n=1}^{\infty }{x}_n$ is one whose partial sums $s_N={\sum }_{n=1}^N{x}_n$ converge to a finite real limit.

If $s_N\to s$, then $x_N=s_N-s_{N-1}\to s-s=0$. This necessary condition is the term test for divergence. Its converse fails: the harmonic series has terms $1/n\to 0$ but diverges. For a direct proof, each block from $n=2^{j-1}+1$ to $2^j$ contains $2^{j-1}$ terms each at least $2^{-j}$, so contributes at least $1/2$. Infinitely many such blocks make the partial sums unbounded.

For the positive sequences in the final request, the inequality makes $y_n$ decreasing and bounded below by zero. The bounded monotone sequence theorem gives $y_n\to l\ge 0$. Summing the decrements givesSuppose $l>0$. Since $y_n\ge l$, this impliesBut clipping preserves divergence of a positive series: if infinitely many $x_n\ge l$, the clipped series has infinitely many terms equal to $l$; otherwise its tail equals the divergent tail of $\sum x_n$. Either case contradicts the displayed bound. Hence the only possible limit isThis proves the minimum-decrement convergence criterion without assuming that the divergent positive sequence $x_n$ itself tends to zero.

The key telescoping series identity isFor the requested mathematical induction, when $n=1$ the proposed sum is $q/(q+1)=1-q!/(q+1)!$. If the identity holds for $n$, adding $c_{n+1}=q!/(q+n)!-q!/(q+n+1)!$ cancels its last term and yields $1-q!/(q+n+1)!$. Thus it holds for every $n\ge 1$. Since $q!/(q+n)!\to 0$ for fixed positive integer $q$, the infinite series has sum $\boxed{1}$.

For the second series, the integral digit condition is $0\le a_n\le n-1$, and in particular $a_1=0$. Its partial sums are increasing and satisfyThus the bounded monotone sequence theorem proves convergence to a real number $S\in [0,1]$. The same telescoping calculation gives, for every $N\ge 1$,If positive digits occur infinitely often, then every tail contains a positive term, so $R_N>0$. If $a_n\le n-2$ occurs infinitely often, then every tail contains a digit smaller than its maximum. Indeed,Consequently the two conditions together give $0<R_N<1$ for every $N$. Were $S=A/B$ a rational number, with positive integer $B$, choose $N\ge B$. Because $B\mid N!$, the number $N!S$ is an integer, as is ${\sum }_{n=1}^N{a}_{n}N!/n!$. Their difference $R_N$ would be an integer strictly between zero and one, a contradiction.

Conversely, if positive digits occur only finitely often, then $S$ is a finite sum of rational numbers. If digits below their maximum occur only finitely often, then $a_n=n-1$ for every $n>N$ for some $N$, givingThese are exactly the two ways the conjunction of infinitude conditions can fail. Hence the irrationality criterion for factorial series is proved in both directions: $S$ is irrational exactly when both kinds of digit occur infinitely often.

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

Fix $x\in (0,1)$ and put $x_0=x$, $x_{n+1}=f(x_n)$. Then $0<x_{n+1}<x_n\le x_0<1$. The bounded monotone sequence theorem gives a limit $\ell \in [0,x_0]$. If $\ell >0$, it lies in the domain, where continuity allows passage to the limit in the recurrence: $\ell =f(\ell )$. That is a fixed point, contrary to $f(\ell )<\ell$. Hence $\boxed{f^n(x)\to 0\text{ for every }x\in (0,1)}$. Continuity is used at a possible positive limiting point, not at the excluded endpoint zero.

No: there need not be even one orbit tending to zero. Partition the domain into the disjoint intervalsand, for the unique $j$ with $x\in I_j$, defineThe left endpoint is strictly below $x$, so $0<f(x)<x$. Moreover $f(x)>1/(j+1)$ and $f(x)\le x\le 1/j$, so $f(x)$ remains in the same interval. For every iterate,Every starting point belongs to some finite-index interval, so this proves the claim for all $x\in (0,1)$. The open left endpoints matter: an orbit approaches a boundary but never crosses it. This is discontinuous trapping of decreasing iterates, not a violation of the bounded monotone sequence theorem; each orbit does converge, just to a positive value where continuity fails.

The arithmetic-geometric mean inequality givesBy induction these inequalities hold at every step, so $(a_n)$ is increasing and bounded above by $b$, while $(b_n)$ is decreasing and bounded below by $a$. The bounded monotone sequence theorem gives limits $\alpha ,\beta$ with $0<a\le \alpha \le \beta \le b$. Taking limits in the arithmetic mean recurrence gives $\beta =(\alpha +\beta )/2$, whence $\boxed{\alpha =\beta }$. Thus both sequences converge to the same positive limit, the arithmetic-geometric mean iteration's limit. No elementary closed formula for that limit is needed.

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.