Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
HINT: Density of $\mathbb{Q}$ in $\mathbb{R}$ –  Don Larynx Sep 29 '13 at 7:55
    
@DonLarynx, not all rational numbers are of the form $S_n-[S_n]$. –  njguliyev Sep 29 '13 at 10:47
1  
By $a\le b$ you mean $a\lt b$, right? –  bof Oct 16 '13 at 12:47
add comment

4 Answers 4

up vote 3 down vote accepted
+50

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.

share|improve this answer
add comment

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 $S_{n+1}-S_n\to0$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
    
This is true, but doesn't prove the "density" result the OP is looking for. –  Callus Oct 16 '13 at 7:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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