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+2 votes
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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 by Veteran (431k points)
recategorized by | 115 views
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

+3 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.

by Boss (18.9k points)
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.
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