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.

  • $\begingroup$ HINT: Density of $\mathbb{Q}$ in $\mathbb{R}$ $\endgroup$ – Don Larynx Sep 29 '13 at 7:55
  • $\begingroup$ @DonLarynx, not all rational numbers are of the form $S_n-[S_n]$. $\endgroup$ – njguliyev Sep 29 '13 at 10:47
  • 3
    $\begingroup$ By $a\le b$ you mean $a\lt b$, right? $\endgroup$ – bof Oct 16 '13 at 12:47
  • $\begingroup$ Some proof or statement of here are not greater than following question. math.stackexchange.com/questions/2062960/… $\endgroup$ – Takahiro Waki Jan 21 '17 at 10:26

I think the direct approach works pretty well here.

Let $N > \max(a^{-1},( b - a ) ^ {-1})$, so that $S_{n+1} - S_{n} < ( b - a )$ and $S_{n+1} - S_{n} < a $ when $n > N$. 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 $n_{0} > N$ be smallest such that $a <S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $. Then $\lfloor S_{n_0} \rfloor = \lfloor S_{n_0 + 1} \rfloor $ and $a > S_{n_0} - \lfloor S_{n_0} \rfloor $ since $S_{n_0 + 1} - S_{n_0} = \frac{1}{n_0 + 1} < a$. Then we also have $S_{n_0+1} - \lfloor S_{n_0 + 1} \rfloor < b$ since $$b > a + \frac{1}{n_0 + 1} > S_{n_0} - \lfloor S_{n_0 } \rfloor + \frac{1}{n_0+1} = S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $$ 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.

  • $\begingroup$ How do you know that there is any $n_0$ with $a < S_{n_0 +1} - \lfloor S_{n_0 +1} \rfloor $? $\endgroup$ – PhoemueX Dec 18 '16 at 10:57
  • $\begingroup$ @PhoemueX The increments are smaller than $b-a$, but the series diverges. It's like there's a puddle a mile long, you start on one side, end on the other, and none of your steps are longer than a mile. You definitely stepped in that puddle. $\endgroup$ – Callus Dec 19 '16 at 14:11

Let $i\geqslant1/(b-a)$ and $k=\lceil S_i\rceil$. Since the sequence $(S_j)_{j\geqslant1}$ is unbounded, some values of this sequence are greater than $k+a$, hence $n=\min\{j\mid S_j\gt k+a\}$ is well defined and finite. Then $S_{n-1}\leqslant k+a\lt S_{n}$ and $n\gt i$ since $S_i\lt k+a$ hence $$ S_{n}=S_{n-1}+1/n\lt k+a+1/i\leqslant k+b. $$ Thus, $k+a\lt S_{n}\lt k+b$, in particular, $\lfloor S_{n}\rfloor=k$ and $a\lt S_{n}-\lfloor S_{n}\rfloor\lt b$.

For every $i$ large enough, this provides some $n\gt i$ such that $a\lt S_{n}-\lfloor S_{n}\rfloor\lt b$. To get infinitely many indexes $n$ such that $a\lt S_{n}-\lfloor S_{n}\rfloor\lt b$, just iterate the construction.

This approach works for every unbounded sequence $(S_n)_n$ such that $S_{n+1}-S_n\to0$.


Since $H_n = \log n +\gamma + O\left(\frac{1}{n}\right)$, it is sufficient to prove the density of the sequence $\{\log n\pmod{1}\}_{n>1}$ in $[0,1]$. But since $\sum_{n=1}^{+\infty}\log\left(\frac{n+1}{n}\right)$ diverges, there must be an element of the sequence in every sub-interval of $[0,1]$: otherwise, take an accumulation point of the sequence (it must exist, because $\{\log n\pmod{1}\}_{n>1}$ is a sequence of distinct real numbers, since all the numbers of the form $e^m$ with $m\in\mathbb{N}_0$ are irrational) and slowly move towards the right ($\log N\rightarrow \log(N+1)\rightarrow\ldots$, with the possibility to choose $N$ arbitrarily big) until falling into the chosen sub-interval.


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 $$S_{2^n}>\dfrac{1}{2}n$$

  • $\begingroup$ This is true, but doesn't prove the "density" result the OP is looking for. $\endgroup$ – Callus Oct 16 '13 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.