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Define what it means for $x_n$ to be bounded.
Prove that a sequence $$x_n=\frac{2n+11}{3n+14}$$ is bounded.

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closed as not constructive by Andrés Caicedo, TMM, Dennis Gulko, muzzlator, Tara B Apr 4 '13 at 18:41

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$x_n$ is bounded if there are constants $m , M$ such that for all $n$, $m \le x_n \le M$. –  Andrew Salmon Apr 4 '13 at 17:05
What have you tried? –  Cameron Buie Apr 4 '13 at 17:06

2 Answers 2

We say a sequence $x_n$ is bounded, if there exists a constant $M \in \mathbb{R}^+$ such that $$\vert x_n \vert < M$$ for all $n \in \mathbb{N}$. Note that $M$ is a constant, i.e., independent of $n$.

Equivalently, a sequence $x_n$ is bounded, if there exist constants $m,M \in \mathbb{R}$ such that $$m < x_n < M$$ for all $n \in \mathbb{N}$. As before $m$ and $M$ are constants, i.e., independent of $n$.

For the sequence $x_n = \dfrac{2n+11}{3n+14}$, clearly $x_n$ is positive. Also, note that $$x_n = 1 - \underbrace{\dfrac{n+3}{3n+14}}_{ \geq 0}$$ Hence, for sure $x_n \in (0,1)$.

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Can't we find $M$ by taking the limit as $n \to \infty$ (i.e. $M = 2/3$)? –  Jacob Apr 4 '13 at 17:25
@Jacob. That need not necessarily be true. For instance, $x_n = \dfrac{2n+1}{3n}$ will give a limiting value of $2/3$. But this doesn't give $M$, in fact this gives the lower bound and not the upper bound. However, it is true that if the limit exists as a real number, then the sequence of real numbers is bounded. However, a sequence can be bounded even if the limit doesn't exist. For instance, $x_n = (-1)^n$. –  user17762 Apr 4 '13 at 17:30
Thanks for your explanation! –  Jacob Apr 4 '13 at 17:33

Note that $x_n$ is positive and clearly that $2n+11<3n+14$ for $n\in\mathbb{N}$ so $$0\leq x_n=\frac{2n+11}{3n+14}\leq1$$ hence $(x_n)$ is bounded.

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