China Mathematical Competition (Complementary Test)

For any $n\in {\mathbb{N}}^{\ast }$, we have $$\begin{array}{rl}{S}_{2^n}& =1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2^n}=1+\frac{1}{2}+\left(\frac{1}{2^1}+\frac{1}{2^2}\right)+\\ & \left(\frac{1}{2^{n-1}}+\frac{1}{2^n}\right)\\ & >1+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\cdots +\left(\frac{1}{2^n}+\cdots +\frac{1}{2^n}\right)\\ & =1+\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{2}>\frac{1}{2}n.\end{array}$$ Let $N_0=\lfloor \frac{1}{b-a}\rfloor +1$, $m=\lfloor S_{N_0}\rfloor +1$. Then $\frac{1}{b-a}<N_0$, $\frac{1}{N_0}<b-a$, and $S_{N_0}<m\le m+a$. Let $N_1=2^{2(m+1)}$. Then $S_{N_1}=S_{2^{2(m+1)}}>m+1\ge m+b$. We claim that there exist $n\in {\mathbb{N}}^{\ast }$ with $N_0<n<N_1$ such that $m+a<S_n<m+b$ (or, in other words, $S_n-[S_n]\in (a,b)$). Otherwise, assuming the claim is false, then there must exist $k>N_0$ such that $S_{k-1}\le m+a$ and $S_k\ge m+b$. Then $S_k-S_{k-1}\ge b-a$. But it contradicts the fact that $$S_k-S_{k-1}=\frac{1}{k}<\frac{1}{N_0}<b-a.$$ Therefore, the claim is true. Furthermore, assume there are only a finite number of positive integers $n_1,\dots ,n_k$ satisfying $$S_{n_j}-[S_{n_j}]\in (a,b)(1\le j\le k).$$ Define $c={\min }_{1\le j\le k}\{S_{n_j}-[S_{n_j}]\}$. Then there exists no $n\in {\mathbb{N}}^{\ast }$ such that $S_n-[S_n]\in (a,c)$. It contradicts the above claim. Therefore, there are infinite terms in the sequence $\{S_n-[S_n]\}$ that are within $(a,b)$. The proof is complete.

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Alternative solution.

For any $n\in {\mathbb{N}}^{\ast }$, we have $$\begin{array}{rl}{S}_{2^n}& =1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2^n}=1+\frac{1}{2}+\left(\frac{1}{2^1}+\frac{1}{2^2}\right)+\\ & \left(\frac{1}{2^{n-1}}+1+\cdots +\frac{1}{2^n}\right)\\ & >1+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{1}{2^2}\right)+\cdots +\left(\frac{1}{2^n}+\cdots +\frac{1}{2^n}\right)\\ & =1+\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{2}>\frac{1}{2}n.\end{array}$$ Therefore, $S_n$ can be larger than any positive number as long as $n$ becomes sufficiently large. Let $N_0=\lfloor \frac{1}{b-a}\rfloor +1$. Then $\frac{1}{N_0}<b-a$, and when $k>N_0$, we have $$S_k-S_{k-1}=\frac{1}{k}<\frac{1}{N_0}<b-a.$$ So for any positive integer $m>S_{N_0}$, there exists $n>N_0$ such that $S_n-m\in (a,b)$, or, in other words, $m+a<S_n<m+b$. Otherwise, there must be $k>N_0$ such that $S_{k-1}\le m+a$ and $S_k\ge m+b$, i.e., $S_k-S_{k-1}\ge b-a$. But it contradicts the fact that $$S_k-S_{k-1}=\frac{1}{k}<\frac{1}{N_0}<b-a.$$ Now let $m_i=[S_{N_0}]+i$ ($i=1,2,3,\dots$). Then there exists $n_i>N_0$ such that $m_i+a<S_{n_i}<m_i+b$, i.e., $S_{n_i}-[S_{n_i}]\in (a,b)$. Therefore, there are infinite terms in the sequence $\{S_n-[S_n]\}$ that are within $(a,b)$. $\square$