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 $0$
 A: You've made things far more complicated than they have to be. You said you know that the sequence $\frac{1}{n}$ converges to zero. How do you know? The definition of the limit of a sequence. For every $\epsilon>0$ there exists $N \in \mathbb{N}$ such that for all $n >N$, $|\frac{1}{n}|<\epsilon$. Specifically, if you choose $N> \frac{1}{\epsilon}$, then if $n >N>\frac{1}{\epsilon}$, you have $\frac{1}{n}<\frac{1}{1/\epsilon}=\epsilon$.
Now try using the definition to show $\frac{1}{\sqrt{n}}$ converges to $0$. That is, let $\epsilon>0$. Now let $N=...?$ and show that if $n>N$, then $\frac{1}{\sqrt{n}}<\epsilon$. You need to figure out what $N$ should be. Try starting with the inequality $\frac{1}{\sqrt{n}}<\epsilon$ and working backward to see how $n$ needs to relate to $\epsilon$.
Edit: Also, the result you stated in your question is false. Consider $c_n=a_nb_n$ with $a_n=\frac{1}{n}$ and $b_n=n$. Then $a_n\to 0$ and $c_n$ is bounded, but $c_n=1 \not\to0$.
A: Not knowing what theorems you have already learned or which you may use, it should be pointed out that a simple $\epsilon$ proof readily shows the limit to be $0$.
Let $\epsilon>0$ and let $n>N=\dfrac{1}{\epsilon^2}$.
Then $\dfrac{1}{n}<\epsilon^2$ thus $\left| \dfrac{1}{\sqrt{n}}-0\right|<\epsilon$.
Therefore,
$$\lim_{n\to\infty}\dfrac{1}{\sqrt{n}}=0$$
A: 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 $a$, $0<a<1$, then we can always find a term $n > \frac{1}{a^2}$ such that $a_n=\frac{1}{\sqrt{n}}<a$, which contradicts $a>0$, so it must be $a=0$.
