Proving the limit of a sequence

Question

I'm having trouble understanding how this works, conceptually.

Let's say I have the simple sequence $u_n = \frac{1}{\sqrt{n}}$. I know that this converges to 0.

I wanna show that, no $\forall \varepsilon>0, \exists N\in \mathbb N$ s.t. $(n > N) \implies \frac{1}{\sqrt{n}} - 0 < \varepsilon$.

So in my head, to prove this for any example, I want to find a N defined as a function of $\varepsilon$. Is this the thought process I should be having?

Thus, I can just say that $\frac{1}{\sqrt{n}} < \varepsilon \implies \frac{1}\varepsilon< \sqrt n \implies \frac{1}{\varepsilon^2}< n $ (since $n$ and $\varepsilon$ are positive.)

And now, what? I think it's proven, right? If we just take $N = \frac{1}\varepsilon$, then the implication works every time? Do I have to show it by saying:

$\forall \varepsilon>0$, if $n > \frac{1}{\varepsilon^2}$, then we have $\frac{1}{\sqrt{\frac{1}\varepsilon}}< \varepsilon \implies \frac{1}{\frac{1}\varepsilon} < \varepsilon \implies \varepsilon < \varepsilon $? But this isnt true,so where exactly am I going wrong? Thanks.

Answer 1

Note that from here we have

$$\frac{1}{\sqrt{n}} < \varepsilon \implies n>\frac{1}{\varepsilon^2}$$

then $\forall \varepsilon>0$ it suffices to choose $n>N>\frac{1}{\varepsilon^2}$ in order to fulfil the definition of limit.

Answer 2

$ \frac{1}\varepsilon< \sqrt n \implies \frac{1}\varepsilon< n $

You appear to be assuming that $\sqrt \epsilon <\epsilon$. But when $\epsilon$<1, taking the square root makes it larger. So for $\epsilon<1$, you need $N>\frac{1}{\epsilon^2}$. If $\epsilon \geq 1$, then you can take N = 2. So you can take $N= max(2, \frac{1}{\epsilon^2})$