# How can I find $\lim_{n\to \infty} a_n$ [duplicate]

Let $$a_n=\left(1-\dfrac{1}{\sqrt2}\right)\dots \left(1-\dfrac{1}{\sqrt{n+1}}\right),n\ge1$$ Then find $\lim_{n\to \infty} a_n$.

Note that $a_n>0$ and $$\lim_{n\to\infty}\log a_n=\sum_{k=1}^\infty \log \left(1-\dfrac{1}{\sqrt{k+1}}\right)$$ is divergent to $-\infty$ by comparaison test with Riemann divergent series: $$\log \left(1-\dfrac{1}{\sqrt{k+1}}\right)\sim_\infty -\dfrac{1}{\sqrt{k}}$$ hence $$\lim_{n\to\infty} a_n=0$$
If you consider $b_n=\ln a_n$, what can you say about the behavior of the sequence $(b_n)_{n\geq 2}$?