For reference, I am in Introduction to Abstract Math, and I have taken Calculus 1, Discrete Math, and Linear Algebra.
I am trying to prove the convergence of $2^{-n}$ as a stepping stone to prove that $(a_n)$ and $(b_n)$ both converge to $\sqrt{2}$, which $(a_n)$ and $(b_n)$ defined as per the following post: Prove that the elements of sequences $(a_n),(b_n)$ are rational numbers such that $a_n<\sqrt{2}<b_n=a_n+2^{-n}$ for all $n \geq 1$ using induction
(I managed to prove that $a_n < \sqrt{2} < b_n = a_n+2^{-n}$)
Here is what I have so far:
Prove that $\frac{1}{2^n}$ converges to $0.$
$\lim\limits_{n\to\infty} \frac{1}{2^n} = 0$ and converges to $0$ if and only if for any $\epsilon > 0$ there is an integer $N>0$ such that if $n > N$ then $\mid a_n \mid < \epsilon$.
Select $N \in \mathbb{Z}^+$ so that $N > \log_2\left(\frac{1}{\epsilon}\right)$.
(I got $\log_2\left(\frac{1}{\epsilon}\right)$ by solving $\mid2^{-n}\mid < \epsilon$ for $n$.)
I am confused as to what my next step would be, where would I go from here?
(I know it would be easier to show in other ways, but my professor wants us to use this way specifically for showing convergence)
Here's how I would prove it. We claim that $2^n>n$ for all integers $n>0$. The base case $2>1$ holds, and if $2^n>n$, then $2^{n+1}=2\cdot 2^n=2^n+2^n>n+n>n+1$.
This shows $\frac{1}{2^n}<\frac1n$ for all $n\in\{1,2,\ldots\}$. Now let $\varepsilon>0$, and (by the Archimedian property of $\Bbb R$) pick a positive integer $N$ such that $N>\frac{1}{\varepsilon}$. That is, $\frac{1}{N}<\varepsilon$. Now for any $n\ge N$, we have $$\left|\frac{1}{2^n}\right|=\frac{1}{2^n}<\frac{1}{n}<\frac{1}{N}<\varepsilon,$$
which shows $\lim\limits_{n\to\infty}\frac{1}{2^n}=0$.
Your method also works though. Working backwards, if we want $\frac{1}{2^n}<\varepsilon$, then $\frac{1}{\varepsilon}<2^n$, so $n>-\log_2(\varepsilon)$. Thus pick $N$ to be any positive integer larger than $-\log_2(\varepsilon)$, etc. The reason I took a different approach is because this method of "solving for $n$ in terms of $\varepsilon$" is in generally very difficult or impossible. It's often much easier to think about how you can bound an expression by something more manageable.