ISI2015-MMA-22

Question

Let $a_n= \bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1- \frac{1}{\sqrt{n+1}} \bigg), \: \: n \geq1$. Then $\underset{n \to \infty}{\lim} a_n$

• A. equals $1$
• B. does not exist
• C. equals $\frac{1}{\sqrt{\pi}}$
• D. equals $0$

Comments

https://math.stackexchange.com/questions/3124615/limiting-value-of-a-sequence-when-n-tends-to-infinity

https://gateoverflow.in/304586/isi-mma-2015

Answer 1

Answer $D$

$$a_n = \Bigg(1-\frac{1}{\sqrt2}\Bigg)\ldots\Bigg(1-\frac{1}{\sqrt{n+1}}\Bigg)$$

Now since,

$$\Bigg(1-\frac{1}{\sqrt{n+1}}\Bigg)\leq\Bigg(1-\frac{1}{n+1}\Bigg)$$

 

$$\implies \Bigg(1-\frac{1}{2}\Bigg)\ldots\Bigg(1-\frac{1}{n+1}\Bigg) = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}\ldots\frac{n}{n+1}$$

 

$$\implies a_n \leq\frac{1}{n+1}$$

Now, substitute $$n \to \infty \implies a_n = 0$$

$\therefore$ $D$ is the correct option.

Comments

One thing to be noted that $a_n\ge 0$ for all $n\in\mathbb{N}$. Hence, by comparison on limits, $\lim_{n\to \infty}a_n \ge 0$. Since $a_n\le0$ as $n\to\infty$ as proved above. Hence, by sandwich theroem. It follows that $a_n=0$ as $n\to\infty$.