# ISI2015-MMA-22

167 views

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
recategorized

1 vote
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.

## Related questions

1
213 views
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
1 vote
The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals $e^{-1}$ $e^{-1/2}$ $e^{-2}$ $1$
The limit $\displaystyle{}\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$
$\displaystyle{}\underset{n \to \infty}{\lim} \frac{1}{n} \bigg( \frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$