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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

1. Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016

Footnotes