Convergence of sequence - Do we have to take these cases?

Question

I want to check the convergence of $\displaystyle{a_n=\frac{2-n}{-4n^2}}$ and give n as a function of $\epsilon$.

For a fixed $\epsilon>0$ we have to find a number $n_{\epsilon}$ from whiich the condition $|a_n- a| < \epsilon$ holds: \begin{equation*}|a_n-a|<\epsilon \iff \left |\frac{2-n}{-4n^2}-0\right |<\epsilon \iff \left |\frac{2-n}{-4n^2}\right |<\epsilon \iff \frac{|2-n|}{4n^2}<\epsilon\iff |2-n|<4n^2\epsilon\iff 4\epsilon n^2-|2-n|>0\end{equation*}

For $2-n>0\Rightarrow n<2$ the $n$ can take only the value $1$.

For $2-n<0$ we have $|2-n|=-(2-n)=-2+n$ so we get $$4\epsilon n^2-n+2>0 \ \ \text{ for } \ \ n<\frac{1- \sqrt{1-32\epsilon}}{8\epsilon} \ \text{ or } \ n>\frac{1+ \sqrt{1-32\epsilon}}{8\epsilon}$$

Are all cases correct?

Answer 1

It is not necessary to use tight bounds, it is often easier to take gross estimates.

For $$\frac{|n-2|}{4n^2}<\epsilon$$ we can use $|n-2|<n$ (which is true for all $n>1$) so that

$$\frac{|n-2|}{4n^2}<\frac1{4n}<\epsilon.$$

Then,

$$n>\frac1{4\epsilon}$$ is fine. (Notice that this bound is just slightly larger than your last expression, and is obtained with less effort.)

Answer 2

Yes , you just have to set:

$$ n_{\epsilon}=\lfloor{\dfrac{1+\sqrt{1-32\epsilon}}{8\epsilon}}+1$$

You can also said its obvious.