= Assumptions of the bounded monotone sequence theorem
Dropping either condition destroys the implication. The increasing <sequence> $a_n=n$ has a <lower bound> one but no <upper bound>, and no finite <limit of a sequence>. For any real threshold $T$, the <Archimedean property> gives an index $N>T$, and monotonicity gives $a_n>T$ for all $n\geq N$. Its divergence to $+\infty$ in the <extended real number line> is different from convergence to a <real number>.
Conversely, $b_n=(-1)^n$ is a <bounded sequence>, with $-1\leq b_n\leq1$, but is not a <monotone sequence> and has no <limit of a sequence>. Its <subsequence> at <even integers> is constantly one, and its <subsequence> at <odd integers> is constantly minus one. These two <subsequential limits> cannot both equal a single <limit of a sequence>. <Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl>[§2.1, Example 2.1.5: divergence of the alternating sequence; §2.1.1, examples immediately after Definition 2.1.9: the increasing sequence n]{ref}
\Image[/monotone-sequence-assumption-counterexamples.png]
{title=Neither monotonicity alone nor boundedness alone ensures a finite <limit of a sequence>}
{description=Left: $a_n=n$ is increasing but unbounded above. Right: $b_n=(-1)^n$ stays in $[-1,1]$ but alternates between two distinct <subsequential limits>. Both plots show finitely many terms; the formulas and the arguments establish the counterexamples for all indices.}
The ambient <real numbers> also matter. An original example using the <floor function> is
$$
q_n=\frac{\lfloor2^n\sqrt2\rfloor}{2^n},\qquad n\geq1.
$$
Every $q_n$ is a <rational number>, $q_n\leq q_{n+1}$ because $\lfloor2x\rfloor\geq2\lfloor x\rfloor$, and
$$
0<\sqrt2-q_n<2^{-n}.
$$
The strict lower inequality uses the <irrationality of the square root of two>. Hence this <bounded sequence> converges to $\sqrt2$ in the <real numbers>, but has no <limit of a sequence> in the <rational numbers>. Convergence to a rational candidate would also be convergence in the <real numbers>, contradicting uniqueness of the <limit of a sequence>. The formulas give $q_1=1$, $q_2=5/4$, $q_3=11/8$, $q_4=11/8$: this counterexample even has a plateau. This illustrates why <completeness of the real numbers> matters: the <rational numbers> lack the <least-upper-bound property> needed to contain the required <supremum>. <Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl>[§1.2.1, Theorem 1.2.1 and Example 1.2.3: least-upper-bound property and the irrational boundary at the square root of two]{ref}
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