Finding the limit of the sequence $(1 - 1/\sqrt 2) \dotsm ( 1 - 1/\sqrt {n+1})$ [duplicate]
Let $$ a_n = \left( 1 - \frac{1}{\sqrt 2} \right) \dotsm \left( 1 - \frac{1}{\sqrt {n+1}} \right), \quad n\geq 1. $$ Then $\lim_{n \to \infty} a_n$
- is $1$,
- is $0$,
- does not exist,
- is $\frac{1}{\sqrt\pi}$.
I tried this using the sandwich theorem as $$ 0 \leq \sum_{r=1}^{n} \frac{-1}{\sqrt r+1} \leq \frac{-n}{\sqrt {n+2}} $$ which goes to zero. I am not sure though.
Solution 1:
Using $$ a_n=\left(1-\frac1{\sqrt2}\right)\cdots\left(1-\frac1{\sqrt{n+1}}\right) $$ then, because $$ \left(1-\frac1{\sqrt{n+1}}\right)\le\left(1-\frac1{n+1}\right) $$ and $$ \begin{align} \left(1-\frac12\right)\cdots\left(1-\frac1{n+1}\right) &=\frac12\cdot\frac23\cdot\frac34\cdots\frac{n}{n+1}\\ &=\frac1{n+1} \end{align} $$ we have $$ a_n\le\frac1{n+1} $$
Solution 2:
$(1-\dfrac{1}{\sqrt {n+1}})^n\geq a_n=(1-\dfrac{1}{\sqrt 2})...(1-\dfrac{1}{\sqrt {n+1}})\geq (1-\dfrac{1}{\sqrt 2})^n$
As $(1-\dfrac{1}{\sqrt 2})<1\implies (1-\dfrac{1}{\sqrt 2})^n\to 0$
Also $(1-\dfrac{1}{\sqrt {n+1}})^n\approx\dfrac{1}{e^{\sqrt n}}\to 0$
Hence $\lim a_n\to 0$