Answer to Question #293684 in Real Analysis for Sarita bartwal

Question

Prove that the sequence {an/n} is convergent where { an} is a bounded sequence

Answer 1

Solution:

Given that $\left\{a_{n}\right\}$  is a bounded sequence.

$\Rightarrow \quad\left|a_{n}\right| \leqslant M$  for all $n \in \mathbb{N}$ , for some $M \in \mathbb{R}^{+}$

$\Rightarrow\left|\frac{a_{n}}{n}\right| \leqslant \frac{m}{n}$

But the sequence $\frac{m}{n}$  converges to 0 .

$\left.\begin{array}{rl} \therefore & \lim _{n \rightarrow \infty}\left|\frac{a_{n}}{n}\right| \leqslant \lim _{n \rightarrow \infty} \frac{m}{n}=0 \\ \Rightarrow & \lim _{n \rightarrow \infty}\left|\frac{a_{n}}{n}\right|=0 \quad\left[\because\left|\frac{a_{n}}{n}\right| \text { is a non negative sequence }\right] \\ \Rightarrow & \lim _{n \rightarrow \infty} \frac{a_{n}}{n}=0 \quad\left[\because-\left|\frac{a_{n}}{n}\right| \leqslant \frac{a_{n}}{n} \leqslant\left|\frac{a_{n}}{n}\right|\right. \text{By sandwich theorem}] \\ \end{array}\right]$

Hence, proved.