Related to USA TST 2000/4 https://artofproblemsolving.com/community/c6h61959p371496. This result can be obtained by using the recursion in post #10 in that thread (we have to prove that the sum of the reciprocals of the coefficients has lim=2).

As stated in the title of the problem we need to show that we can choose $n$, such that: $$\sum_{i=0}^{n}\frac{1}{{n \choose i}} < 2.022$$Notice that for each $i$, $1 < i < n$, we have that: $$\frac{1}{{n \choose 2}} \geq \frac{1}{{n \choose i}}$$thus we have that: $$LHS = 2+\sum_{i=1}^{n-1}\frac{1}{{n \choose i}} \leq 2+(n-2)\frac{2}{n(n-1)} < 2.022$$which implies that we must show that, there exists an $n$ such that: $$\frac{n-2}{n(n-1)} < 0.011$$but notice that we have that: $$\lim_{n \rightarrow \infty} \frac{n-2}{n(n-1)} = 0$$Which means that as $n$ gets bigger the fraction gets smaller and smaller, thus there exists an $n$ so that the sum is less than $2.022$

We have to prove that \[\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{1}{{n \choose k}}} = 2\] First, considering \[{T_n} = \frac{1}{{\left( \begin{array}{c} n\\ 0 \end{array} \right)}} + \frac{2}{{\left( \begin{array}{c} n\\ 1 \end{array} \right)}} + \frac{3}{{\left( \begin{array}{c} n\\ 2 \end{array} \right)}} + ... + \frac{{n + 1}}{{\left( \begin{array}{c} n\\ n \end{array} \right)}} = \sum\limits_{k = 0}^n {\frac{{k + 1}}{{\left( \begin{array}{c} n\\ k \end{array} \right)}}} \text{ (1)}\]Since $\left( \begin{array}{c} n\\ k \end{array} \right) = \left( \begin{array}{c} n\\ n - k \end{array} \right)$, we have \[{T_n} = \frac{{n + 1}}{{\left( \begin{array}{c} n\\ 0 \end{array} \right)}} + \frac{n}{{\left( \begin{array}{c} n\\ 1 \end{array} \right)}} + \frac{{n - 1}}{{\left( \begin{array}{c} n\\ 2 \end{array} \right)}} + ... + \frac{1}{{\left( \begin{array}{c} n\\ n \end{array} \right)}} \text{ (2)}\] Add (1) and (2) we have $$ 2T_n=(n+2)a_n $$Also notice that $\frac{{k + 1}}{{\left( \begin{array}{c} n\\ k \end{array} \right)}} = \frac{{n + 1}}{{\left( \begin{array}{c} n + 1\\ k + 1 \end{array} \right)}}$, we have $$ T_n=(n+1)(a_{n+1}-1) $$Thus we have a recursion here $$ a_n=\frac{n+1}{2n}a_{n-1}+1 $$ From the recursion, it's easy to see that $$ a_{n+1}<a_n \Leftrightarrow a_n>2+\frac{2}{n} \text{ , which is always right by induction} $$Thus $(a_n)$ is strictly decreasing, and is lower bounded, which means $(a_n)$ converges From the recursion, we conclude that \[\mathop {\lim }\limits_{n \to \infty } {a_n} = 2\]

It suffices to prove $\sum_{i=0}^n \binom{n}{i}^{-1} < 2$.$023$ for some $n$. Equivalently, we want $\sum_{i=1}^{n-1} \binom{n}{i}^{-1} < 0$.$023$. Each of the binomial coefficients except the two ending ones (these are $n-4$ in number) is at least as large as $\binom{n}{2} = \frac{n(n-1)}{2}$, so the main sum does not exceed $\frac{2}{n} + \frac{2(n-4)}{n(n-1)} < \frac{4}{n}$. The latter can be arbitrarily small for sufficiently large $n$ (in particular, less than $0$.$02 < 0$.$023$ for $n \geq 200$).