Answer to Question #159884 in Calculus for sai

Question

Prove that a non-decreasing (resp. non-increasing) sequence which is not bounded above (resp. bounded below) diverges to +∞ (resp. to −∞).

Answer 1

1) Let $\{a_n\}_{n=1}^{+\infty}$  is a non-decreasing sequence which is not bounded above, that is, $\sup\limits_{n}a_n = +\infty$. This means that for all M>0 there exists $N\in\mathbb{N}$ such that $a_N>M$. As the sequence $\{a_n\}_{n=1}^{+\infty}$ is a non-decreasing we have $a_n>M$ for all $n>N.$

Finally: for all M>0 there exists $N\in\mathbb{N}$ such that for all $n>N$ $a_n>M$. Therefore, $\{a_n\}_{n=1}^{+\infty}$diverges to +∞ by the definition.

2) Let $\{a_n\}_{n=1}^{+\infty}$ is a non-inreasing sequence which is not bounded below, that is, $\inf\limits_{n}a_n = -\infty$. This means that for all M<0 there exists $N\in\mathbb{N}$ such that $a_N<M$. As the sequence $\{a_n\}_{n=1}^{+\infty}$ is a non-increasing we have $a_n<M$ for all $n>N.$

Finally: for all M<0 there exists $N\in\mathbb{N}$ such that for all $n>N$ $a_n<M$. Therefore, $\{a_n\}_{n=1}^{+\infty}$diverges to -∞ by the definition.