Proving $\frac{1}{\sqrt{n}}$ converges to $0$ So $\frac{1}{\sqrt{n}}=\frac{\sqrt{n}}{n}=\frac{1}{n}\sqrt{n}$ Let $a_n=\frac{1}{n}$ and $b_n=\sqrt{n}$ so that $c_n=a_nb_n=\frac{1}{n}\sqrt{n}$ $c_n$ convers to $0$ iff $a_n->0$ and $c_n$ is bounded. Okay so we know that $a_n->0$ (converges to $0$). To prove $c_n$ is bounded : $c_n$ is bounded from below by $0$ and we can't find an above bound for $c_n$ By definition $c_n$ has to be bounded for $b_n$ to converge to $0$ But If $c_n$ was bounded by monotonic sequence property it will converge to something else. I'm stuck here, Am I misunderstanding something here? does $c_n$ need to converge for the multiplication of sequences ($b_n$) converge to $0$

The sequence $a_n=\frac{1}{\sqrt{n}}$ is monotonic descending and bounded, thus converging. I.e. $0<\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}\leq 1$

Now, if we assume that it converges to a number $a$, $0 \frac{1}{a^2}$ such that $a_n=\frac{1}{\sqrt{n}}0$, so it must be $a=0$.