Lets use the proof by contradiction. $0.001 = \epsilon$ Claim 1. $n\ge 2020$ then $| a_n(2+ \epsilon)^n |\ge \sum_{k=0} ^{n-1} |a_k(2+ \epsilon)^k|$ if not. there exists some nonnegative numbers which satisfies $ \sum_{i=0}^{n-1}{c_i}=1, c_i | a_n(2+ \epsilon)^n | \leq |a_i(2+ \epsilon)^i |$ (If we take $c_0 ,...,c_{n-2}$ which satisfies the equality, then $c_{n-1}$ is well-defined in that range because of the claim) It leads that for real numbers whose absolute values not smaller than $2+\epsilon$, inequality $c_i |a_n x^n| \leq |a_i x^i|$ holds. From that inequality, we can easily know that $| a_n x^n | - \sum_{k=0} ^{n-1} |a_k x^k|=0$ doesn't have a solution whose absolute value is not smaller than $2+\epsilon$ Contradiction. Claim 2.There leads a scale error in inequality by Claim 1. By telescoping the inequalites in weight of power of 2, $ \sum _{k=2020} ^n {|a_k(2+ \epsilon)^k|\cdot {\frac{1}{2^k}}} \ge \sum _{k=2020} ^n{\sum_{i=0} ^{k-1} |a_i(2+\epsilon)^i| \cdot{ \frac{1}{2^k}}}$ If we look at the coefficients, they are over 0 if between $|a_{2020} |$ and $|a_{n-1}|$ The coefficients of $|a_i |(i\leq 2020)$ is higher than $-C$ The coefficient of $|a_n |$ is $(1+ \frac{\epsilon}{2})^n$ Because $|a_n |\ge 1$, it makes contradiction for enough big $n$ Thanks for CHLORG1 for finding some type errors(I'm not used to write with this LaTeX)

Can you explain more about this line? Thank you.

Oh. I was too impatient in writing the solution. There would be some edits after this post

lkjin7932 wrote: It leads that for real numbers whose absolute values not smaller than $2+\epsilon$, inequality $c_i |a_n x^n| \leq |a_i x^i|$ holds. I don't think this is true. Okay, I see where is the typo. The claim should be the opposite direction.

No. Suppose such a sequence existed. For $n \geq 2020$, we should have $$a_n=\frac{a_{n-1}r^{n-1}+\cdots+a_0}{r^n} \implies |a_n| \leq \frac{|a_{n-1}|}{|r|}+\frac{|a_{n-2}|}{|r|^2}+\cdots+\frac{|a_0|}{|r|^n}$$for some real $r$ with absolute value greater than $2.001$. Since $|r|^{-1}+|r|^{-2}+\cdots=\tfrac{1}{|r|-1}<1$, it immediately follows that $|a_n|$ is bounded above by $\max\{a_0,\ldots,a_{2019}\}$. Now suppose that $|a_i|$ is bounded above by some positive integer constant $C$ for all $i\geq N$. For all $n>N$, we have $$|a_n| \leq \frac{|a_n|}{|r|}+\cdots+\frac{|a_N|}{r^{n-N}}+\frac{|a_{N-1}|r^{N-1}+\cdots+|a_0|}{r^n} \leq \frac{C}{|r|-1}+\frac{|a_{N-1}|r^{N-1}+\cdots+|a_0|}{r^n}.$$By sending $n \to \infty$, the second term vanishes, so we find that for every $\varepsilon>0$ it is true that $|a_n| \leq \tfrac{C}{1.001}+\varepsilon$ when $n$ is large enough. Thus we get that $\tfrac{C}{1.0001}$ is also an upper bound by selecting an appropriate $\varepsilon$. Since the terms in the sequence are integers, $\lfloor \tfrac{C}{1.0001}\rfloor \leq C-1$ is also an upper bound, so by repeating this many times it follows that $|a_i|$ is bounded above by $0$ for sufficiently large $i$: absurd. $\blacksquare$

Let $x_n>2.001$ be the root of $a_n x^n+a_{n-1}x^{n-1}+\cdots +a_0=0.$ $$|a_nx_n^{n}|\le\sum_{0\le k<n}|a_kx_n^k|$$$$\Longrightarrow |a_n\cdot 2.001^n|\le\sum_{0\le k<n}|a_k\cdot 2.001^k|.$$$$\Longrightarrow |a_n|\cdot 2.001^n<2^{n-2021}\cdot\sum_{0\le k\le 2020}|a_k\cdot 2.001^k|,\quad\forall n\ge 2021$$by induction. This immediately gives contradiction since upper bound of $|a_n|\to 0.\Box$