A problem on calculating a limit as $n \to \infty$

Question

What is the value of

$$\lim _{n\to\infty} (1-\frac{1}{\sqrt2}) \dots (1-\frac{1}{\sqrt{n+1}})$$

Answer 1

Hint:

$\log(a_1a_2 \cdots a_n)=\sum_1^n\log a_k$

Answer 2

The product $$a_n:=\left( 1-\frac1{\sqrt2}\right)\left( 1-\frac1{\sqrt3}\right)\cdots\left( 1-\frac1{\sqrt{n+1}}\right) $$ has several factors $1-\frac1{\sqrt i}$ with $1-\frac1{k+1}\le 1-\frac1{\sqrt i}<1-\frac1k$, namely those corresponding to $i=k^2+1,k^2+2,\ldots, (k+1)^2$. Thus for each $k$, $1\le k\le \sqrt{n+1}$, the product $a_n$ has $2k+1$ such factors. Since all remaining factors are $<1$ (and all factors are positive), we find $$ a_n\le \prod_{k=1}^{\lfloor\sqrt {n+1}\rfloor}\left(1-\frac1k\right)^{2k+1}.$$ Because $\left(1-\frac1k\right)^k\to e^{-1}$ and $\left(1-\frac1k\right)^{2k}\to e^{-2}<\frac14$ as $k\to\infty$, we have $\left(1-\frac1k\right)^{2k+1}<\frac14$ for all sufficiently big $k$, say for $k>N$. Then for $n>N^2$, we have $$a_n<\prod_{k=1}^N\left(1-\frac1k\right)^{2k+1}\;\cdot\; \left(\frac14\right)^{\lfloor\sqrt{n+1}\rfloor -N}$$ hence $a_n\to 0$.