How to prove that $a
Let $S_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, where $n$ is a positive integer. Prove that for any real numbers $a,b,0\le a\le b\le 1$, there exist infinite many $n\in\mathbb{N}$ such that
  $$a<S_n-[S_n]<b$$
  where $[x]$ represents the largest integer not exceeding $x$.
This problem is from China 2012 China Second Round (High school math competition) competition last problem, I think this problem has more nice methods, maybe using analytic methods.

since
$$S_{2^n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^n}>1+\dfrac{1}{2}+\left(\dfrac{1}{2^2}+\dfrac{1}{2^2}\right)+\cdots+\left(\dfrac{1}{2^n}+\cdots+\dfrac{1}{2^n}\right)$$
so
$$S_{2^n}>\dfrac{1}{2}n$$
