How to prove that $a<S_n-[S_n]<b$ infinitely often 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
  $$a&lt;S_n-[S_n]&lt;b$$
  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.
I think the direct approach works pretty well here.  

Let $N &gt; \max(a^{-1},( b - a ) ^ {-1})$, so that $S_{n+1} - S_{n} &lt; ( b - a )$ and $S_{n+1} - S_{n} &lt; a $ when $n &gt; N$.  Now suppose there are only finitely many $n$ with the desired property, and increase $N$ so that it's larger than the greatest index for which the property holds.  Finally, let $n_{0} &gt; N$ be smallest such that $a &lt;S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $.  Then $\lfloor S_{n_0} \rfloor = \lfloor S_{n_0 + 1} \rfloor $ and $a &gt; S_{n_0} - \lfloor S_{n_0} \rfloor $ since $S_{n_0 + 1} - S_{n_0} = \frac{1}{n_0 + 1} &lt; a$.  Then we also have $S_{n_0+1} - \lfloor S_{n_0 + 1} \rfloor &lt; b$ since $$b &gt; a + \frac{1}{n_0 + 1} &gt; S_{n_0} - \lfloor S_{n_0 } \rfloor  + \frac{1}{n_0+1} = S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $$  Which means we've found another index where the property holds, contradicting the assumption that $N$ is larger than the largest index for which the property holds.
since
$$S_{2^n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^n}&gt;1+\dfrac{1}{2}+\left(\dfrac{1}{2^2}+\dfrac{1}{2^2}\right)+\cdots+\left(\dfrac{1}{2^n}+\cdots+\dfrac{1}{2^n}\right)$$
so
$$S_{2^n}&gt;\dfrac{1}{2}n$$