Question 54
Let $\alpha\in(0,1)$. Prove that there is a sequence $(\varepsilon_n)_{n\geq1}$, each of whose terms is $0$ or $1$, such that the sequence $(s_n)_{n\geq1}$ defined by
$$
s_n=\frac{\varepsilon_1}{n(n+1)}+\frac{\varepsilon_2}{(n+1)(n+2)}+\cdots+\frac{\varepsilon_n}{(2n-1)2n}
$$
satisfies
$$
0\leq\alpha-2ns_n\leq\frac{2}{n+1}
$$
for every $n\geq2$.
Notation note: the source writes $(\varepsilon_n)_{n\gt1}$ but uses $\varepsilon_1$ in its defining formula. The sequence index has been restored to $n\geq1$ so that the formula is defined.
$$
s_n=\frac{\varepsilon_1}{n(n+1)}+\frac{\varepsilon_2}{(n+1)(n+2)}+\cdots+\frac{\varepsilon_n}{(2n-1)2n}
$$
satisfies
$$
0\leq\alpha-2ns_n\leq\frac{2}{n+1}
$$
for every $n\geq2$.
Notation note: the source writes $(\varepsilon_n)_{n\gt1}$ but uses $\varepsilon_1$ in its defining formula. The sequence index has been restored to $n\geq1$ so that the formula is defined.