ISI2014-DCG-2

2 votes
243 views

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

1. equals $1$
2. does not exist
3. equals $\frac{1}{\sqrt{\pi}}$
4. equals $0$
in Calculus
recategorized
0
yes, it is a consecutive multiplication, if there will be consecutive summation , then ans will be 1.

But here that is not the case
0
So, this is multiplication, confirmed??

1 Answer

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.

edited by
0
Fully the perfect one.
0
BTW writing $n=\infty$ is not purely mathematical. Rather writing $n \to \infty$ is of worthy.
0
Thanks
1
Yes, right.

Corrected now.

Related questions

4 votes
4 answers
1
346 views
$\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$
3 votes
2 answers
2
250 views
$\underset{n \to \infty}{\lim} \dfrac{1}{n} \bigg( \dfrac{n}{n+1} + \dfrac{n}{n+2} + \cdots + \dfrac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$
2 votes
2 answers
3
145 views
$\underset{x \to 2}{\lim} \dfrac{1}{1+e^{\frac{1}{x-2}}}$ is $0$ $1/2$ $1$ non-existent
0 votes
0 answers
4
150 views
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$