Answer $D$
$$a_n = \Bigg(1-\frac{1}{\sqrt2}\Bigg)\ldots\Bigg(1-\frac{1}{\sqrt{n+1}}\Bigg)$$
Now since,
$$\Bigg(1-\frac{1}{\sqrt{n+1}}\Bigg)\leq\Bigg(1-\frac{1}{n+1}\Bigg)$$
$$\implies \Bigg(1-\frac{1}{2}\Bigg)\ldots\Bigg(1-\frac{1}{n+1}\Bigg) = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}\ldots\frac{n}{n+1}$$
$$\implies a_n \leq\frac{1}{n+1}$$
Now, substitute $$n \to \infty \implies a_n = 0$$
$\therefore$ $D$ is the correct option.