Show that the $\lim_{n\to+∞} 1/\sqrt{n} = 0$
So basically my assistant told me i have to show first of all that x→√x is a strictly increasing function. So i did $√x_n < √x_{n+1}$ so $√x_n - √x_{n+1} < 0$. I thought that would be enough. Do i have to show also that $x_n < x_{n+1}$ also. I basically did the conjugate multiplication where $\frac{(√x_n - √x_{n+1})(√x_n + √x_{n+1})}{(√x_n + √x_{n+1})}$ gives us $\frac{x_n < x_{n+1}}{(√x_n + √x_{n+1})} < 0$ which shows that $x_n < x_{n+1}$. Basically my first question is, do i have show that if √x is increasing, then x is also increasing? Or do i show that if x is increasing, then √x is increasing? Or does one of them not imply the other? Is this how i would show it or should i show it differently? Because my assistant hinted that i should multiply using conjugate.
Moreover, concerning the epsilon. I know the definition of a limit etc. I know that for all ε > 0, there is an $N_ε$ such that for all $n≥N_ε$, then $|x_n−x|< ε$. So here, do i define x as 0? Because i just wrote, there exists $N_ε$ such that for all $n≥N_ε$, $|\frac{1}{√n}$ - 0 | < ε. So let $N_ε$ > $\frac{1}{ε^2}$ so then $n$ > $\frac{1}{ε^2}$ iff √n > $\frac{1}{ε}$ iff $\frac{1}{√n}$ < ε which implies that $|\frac{1}{√n}$ - 0 | < ε. Are the steps wrong in this? Can someone tell me the right order of how to go about this?
Thanks!
n -> ∞ : For any ε >0 , there is a $N_ε$ such as : n ≥ $N_ε$ => $n ≥ \frac{1}{ε^2}$
Let $U_n = \frac{1}{√n}$
Then, For any ε >0 : n ≥ $N_ε$ => $0 < U_n \leq ε$
Hence : $(U_n)$ -> 0 when n -> ∞