If $a_n=\left(1-\frac{1}{\sqrt{2}}\right)\ldots\left(1-\frac{1}{\sqrt{n+1}}\right)$ then $\lim_{n\to\infty}a_n=?$

Question

I have an objective type question:-
If $$a_n=\left(1-\frac{1}{\sqrt{2}}\right)\ldots\left(1-\frac{1}{\sqrt{n+1}}\right)$$ then $\lim_{n\to\infty}a_n=?$:-
A)$0$
B)limit does not exist
C)$\frac{1}{\sqrt\pi}$
D)1
My approach :-As the product contains $1/\sqrt2$ so the overall product would be irrational,and as product is converging also so by option elimination answer would be $$C) \frac{1}{\sqrt \pi}$$,as it is the only irrational number in options.
Now i want to ask is that is my solution right? If it is right then what is the proper method to actually solve the question? but if it is wrong then what is the right solution?
Suddenly a doubt is also arising that whether 0 is an irrational number?

Answer 1

A simple estimate yields $$0 < a_n = \left( 1- \frac{1}{\sqrt{2}} \right) \cdots \left( 1- \frac{1}{\sqrt{n+1}} \right) < \left( 1- \frac{1}{\sqrt{n+1}} \right)^n = b_n.$$ The series$$\sum \limits_{n=1}^\infty b_n$$ converges by the comparison test. Thus $b_n \to 0$ and $a_n \to 0$.

Answer 2

\begin{align}\lim_{n\to\infty}a_n&=\left(1-\frac1{\sqrt2}\right)\ldots\left(1-\frac1{\sqrt{n+1}}\right)\le\lim_{n\to\infty}\left(1-\frac1{\sqrt{n+1}}\right)^{n+1}\!\!\Big/\left(1-\frac1{\sqrt{n+1}}\right)\\&=\lim_{x\to\infty}\left(1-\frac1{x}\right)^{x^2}\!\!\Big/\ 1=\lim_{x\to\infty}\left(\left(1-\frac1{x}\right)^x\right)^x=\left(\frac1e\right)^{\infty}=0\end{align}

Answer 3

The real issue here is that the natural logarithm of $(1-\frac 1{\sqrt n})$ is roughly $-\frac 1{\sqrt n}$ (in the sense that the ratio goes to $1$). So the sum of the logarithms of these terms is roughly the partial sum of $-\sum \frac 1{\sqrt n}$. Since the latter diverges, the logarithms go to $-\infty$, so the limit of $a_n$ is $0$.

Answer 4

$\displaystyle \ln a_n = \sum_{i=2}^{n+1} \ln(1-\frac{1}{\sqrt{i}})$ and $\displaystyle \ln(1-\frac{1}{\sqrt{i}})\sim -\frac{1}{\sqrt{i}}$

Thefore $\sum_{i=2}^{n+1} \ln(1-\frac{1}{\sqrt{i}})$ diverges to $-\infty$

Hence $\ln a_n\to -\infty$

Hence $a_n\to 0$