Let $\alpha$ be a real number in the interval $(0,1).$ Prove that there exists a sequence $(\varepsilon_n)_{n\geq 1}$ where each term is either $0$ or $1$ such that the sequence $(s_n)_{n\geq 1}$ \[s_n=\frac{\varepsilon_1}{n(n+1)}+\frac{\varepsilon_2}{(n+1)(n+2)}+...+\frac{\varepsilon_n}{(2n-1)2n}\]verifies the inequality \[0\leq \alpha-2ns_n\leq\frac{2}{n+1}\]for any $n\geq 2.$
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Tags: algebra, Sequence, romania, Romanian TST, TST
02.07.2021 17:29
Just choose $\varepsilon_{n}$ for $n\in\mathbb{N}$, so that $\sum_{i=1}^{n}\varepsilon_{i}=\lfloor\alpha n\rfloor$ for all $n\in\mathbb{N}$. Now let us prove this works. We will say $(a_i)_{i\geq1}\geq(b_i)_{i\geq1}$ if $\sum_{i=1}^{n}a_i\geq\sum_{i=1}^{n}b_i$ for all natural $n$. It is easy to prove that if $(a_i)_{i\geq1}\geq(b_i)_{i\geq1}$ then $\sum_{i=1}^{n}\frac{a_{i}}{(n+i-1)(n+i)}\geq\sum_{i=1}^{n}\frac{b_{i}}{(n+i-1)(n+i)}$ for all natural $n$. Now let $(\delta_i)_{i\geq1}$ be chosen so that $\sum_{i=1}^{n}\delta_{i}=\lfloor\alpha n\rfloor+1$ for all $n$. Then $(\delta_i)_{i\geq1}\geq(\alpha)_{i\geq1}\geq(\varepsilon_i)_{i\geq1}$. Now notice that $2n\sum_{i=1}^{n}\frac{\alpha}{(n+i-1)(n+i)}=2n\sum_{i=1}^{n}\alpha(\frac{1}{n+i-1}-\frac{1}{n+i})=2n\alpha(\frac{1}{n}-\frac{1}{2n})=\alpha$. So $\alpha-2ns_n\geq0$. Now it's enough to prove that $2n\sum_{i=1}^{n}\frac{\delta_i}{(n+i-1)(n+i)} - 2n\sum_{i=1}^{n}\frac{\varepsilon_i}{(n+i-1)(n+i)}=\frac{2}{n+1}$. Notice that $\varepsilon_1=0$, $\delta_1=1$ and $\varepsilon_{i}=\delta_{i}$ for $i\geq2$. Then this difference is equal to $2n\frac{1}{n(n+1)}=\frac{2}{n+1}$, done.
02.07.2021 22:19
Very nice
03.04.2022 15:20
this problem can also use Abel to solve.