I am looking for help to see if I understand proofs by epsilon-delta correctly. I want to prove that the sequence $\{b-\frac{33}{2n^2}\}$ converges to b. Where b is a real number.
I want to prove it with this definition of convergence $\forall\epsilon >0, \exists M>0$ s.t. if $n\geq M$ then $|a_n-a|<\epsilon$.
My work
I start by looking at
$$|b-\frac{33}{2n^2}-b|<\epsilon$$
To determine $\epsilon$, I start out by simplifying the expression on the left side
$$|b-\frac{33}{2n^2}-b|<\epsilon\Leftrightarrow |-\frac{33}{2n^2}|<\epsilon$$ Since the b's eliminate eachother I remove those. I can also remove the negative because of the absolutes, thus I have got the following $$|\frac{33}{2n^2}|<\epsilon$$ Since $2n^2$ can never be negative I can safely remove the absolutes which leaves me with (What if the sequence could be negative, should I look at it casewise?) $$\frac{33}{2n^2}<\epsilon\Leftrightarrow n>\frac{\sqrt{33}}{\sqrt{2}\cdot\sqrt{\epsilon}}$$ I can therefore let $M=\frac{\sqrt{33}}{\sqrt{2}\cdot\sqrt{\epsilon}}$ s.t. $n>M$ and therefore I can conclude that whatever the value of $\epsilon$, I can pick a M that is smaller or equal to n and thus this sequence converges to b.
Have I done it correctly, and where can I improve? I want to be as precise as possible.
Thanks!
I think it suffices to show that $\frac{33}{2n^2} \to 0$.
Fix arbitrary $\epsilon > 0$. If you choose $M > \sqrt{\frac{33}{2\epsilon}}$, you have the following:
$\frac{33}{2M^2} < \frac{33}{2(\sqrt{\frac{33}{2\epsilon}})^2} = \frac{33}{33/\epsilon} = \epsilon$. Since you can make $\epsilon$ arbitrarily small and always find a suitable $M$, then $\frac{33}{2n^2} \to 0$ as $n \to \infty$.
Since $\frac{33}{2n^2} \to 0$ as $n \to \infty$ and $b$ is constant, then I think you have proven that the sequence converges to $b$.