2 votes

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

- equals $1$
- does not exist
- equals $\frac{1}{\sqrt{\pi}}$
- equals $0$

6 votes

__ Answer:__ $\mathbf D$

**Explanation:**

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

Now,

$\because \left(1-\frac{1}{\sqrt{n+1}}\right)\leq\left(1-\frac{1}{n+1}\right)$

$\Rightarrow \left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{n+1}\right) = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}\ldots\frac{n}{n+1}$

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

On substituting $n \to \infty \\\Rightarrow a_n = 0$

$\therefore \mathbf D$ is the correct answer.