prove that there exists no integer in $\pm\frac1k \pm \frac1{k+1}\pm\frac1{k+2}...\pm\frac1{k+n} $

Question

Prove that there exists no integer among the $2^{n+1}$ numbers

$$\pm\frac1k \pm \frac1{k+1}\pm\frac1{k+2}...\pm\frac1{k+n} $$

That is,

$$\mathbb{Z} \cap \left\lbrace \frac{\delta_0}{k} + \frac{\delta_1}{k + 1} + \ldots + \frac{\delta_n}{k+n} : \delta_0, \ldots, \delta_n \in \lbrace -1, 1 \rbrace \right\rbrace = \emptyset$$

This is a discrete maths homework question with another part preceding it which wants us to prove that "in any block of consecutive positive integers there is a unique integer divisible by a higher power of $2$ than any of the others". This I could prove (by negation) so you can use this statement to answer the question.

Any suggestions of approaches to the problem are welcome.

Answer 1

Let $m$ be the exponent of this maximum power of 2, so $2^m$ divides exactly one denominator.

Consider the lcm of the denominators $d$. $2^m|d$. When all the fractions are written in the form $c/d$, the one whose denominator is divisible by $2^m$ will have an odd numerator and all the others will have an even numerator. Therefore their sum will be odd, so the resulting fraction will have an odd numerator and even denominator and so can not be an integer.

Answer 2

Factor all the $n+1$ integers $k,k+1,\cdots,k+n$ and extract the factor $2$ all together. We have $$i=2^{\lambda_i}p_i,$$ where $i=k,k+1,\cdots,k+n.$

Apparently, $\lambda_i \geq 0$ and $2 \nmid p_i.$ Denote $\lambda=\max\{\lambda_i\}.$ Then $\lambda>0$ when $n\geq 1.$ Morover, we may prove there exists only one $i$ such that $i=2^\lambda p_i$. Otherwise, if $i=2^\lambda p_i$ and $j=2^\lambda p_j$ but $i \neq j$, then there exists another even integer between $p_i $ and $p_j$(since $p_i$ and $p_i$ are two different odd integers), which implies there exists a $l$ between $i$ and $j$ such that $l=2^{\lambda+1}p_l.$ This contradicts the assumption that $\lambda$ is the largest.

Now, assume $m=2^\lambda p_m.$ Multiply $$\pm\frac{1}{k}\pm \frac{1}{k+1}\pm \cdots \pm \frac{1}{k+n}$$

by $X=2^{\lambda-1}p_{k}p_{k+1}\cdots p_{k+n}$. Apparently, $M$ could be divided by almost every denominator except for $m$. Thus, we may write $$\left(\pm\frac{1}{k}\pm \frac{1}{k+1}\pm \cdots \pm \frac{1}{k+n}\right)X=Y\pm\frac{p_kp_{k+1}\cdots p_{k+n}}{2 p_m},\tag1$$ where $Y$ is an integer. Notice that $p_k,p_{k+1},\cdots,p_{k+n}$ are all odd integers. Hence, the product of them except for $p_m$ could not be divided by $2$. Therefore, the right-hand side of $(1)$ is not an integer, from which we may complete the proof.