◀ ▲ ▶Branches / Analysis / Application: Application of Monotonic Convergence
The theorem of monotonic convergence is very useful to prove that a given sequence is convergent, without having to find the actual limit of the convergent sequence. The following example demonstrates this.# Application: Application of Monotonic Convergence
(related to Theorem: Every Bounded Monotonic Sequence Is Convergent)
 Consider a sequence $(a_n)_{n\in\mathbb N}$ defined by
$$a_0:=2,\quad\quad a_{n+1}:=\sqrt{a_n + 12}$$
for all $n\ge 0.$
 Then, $a_0=2 < \sqrt{14} < a_1.$
 Suppose that for some fixed $n\ge 0$ we have $a_n < a_{n+1}$.
 Then it follows from $a_n + 12 < a_{n+1} + 12$ that $\sqrt{a_n + 12} < \sqrt{a_{n+1} + 12}$, i.e. that $a_{n+1} < a_{n+2}$.
 By induction, this means that $a_n < a_{n+1}$ for all $n\ge 0.$
 Thus, the sequence $(a_n)_{n\in\mathbb N}$ is strictly monotonically increasing.
 Similarly, $a_0=2 < 4.$
 Suppose that for some fixed $n\ge 0$ we have $a_n < 4$.
 Then it follows from $a_{n+1}=\sqrt{a_n + 12} < \sqrt{4 + 12} = \sqrt{16} = 4.$
 By induction, this means that $a_n < 4$ for all $n\ge 0.$
 Thus, the sequence $(a_n)_{n\in\mathbb N}$ is bounded from above.
 Now, the theorem of monotonic convergence tells us that the sequence $(a_n)_{n\in\mathbb N}$ is convergent, even though we haven't calculated its limit.^{1}
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References
Bibliography
 Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016
Footnotes