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

  1. equals $1$
  2. does not exist
  3. equals $\frac{1}{\sqrt{\pi}}$
  4. equals $0$
in Calculus by Veteran (425k points)
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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.
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