Question #293684

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

Expert's answer

Solution:

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

⇒∣an∣⩽M\Rightarrow \quad\left|a_{n}\right| \leqslant M  for all n∈Nn \in \mathbb{N} , for some M∈R+M \in \mathbb{R}^{+}

⇒∣ann∣⩽mn\Rightarrow\left|\frac{a_{n}}{n}\right| \leqslant \frac{m}{n}

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

∴lim⁡n→∞∣ann∣⩽lim⁡n→∞mn=0⇒lim⁡n→∞∣ann∣=0[∵∣ann∣ is a non negative sequence ]⇒lim⁡n→∞ann=0[∵−∣ann∣⩽ann⩽∣ann∣By sandwich theorem]]\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.


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