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Source. The displayed theorem (1) on p. 1 of the eight-page offprint from Matematikai és Fizikai Lapok 39 (1932); the lemma (2), the notation (3) and the deduction (4) on p. 2; the proof of the lemma on pp. 2--6; the cases $d=1$, $d$ odd and $d=2$ on pp. 6--7; a German summary on p. 8. The scan's OCR layer garbles the formulas; everything below was read on the rendered page images.

Read depth. Claims checked: theorem (1) and lemma (2) were read clause by clause on the rendered page images; the proof (pp. 2--7) was read for structure only and is recorded below as a sketch, not verified.

Statement

Let $a$, $d$, $n$ be arbitrary positive integers. Then

$$\frac1a+\frac1{a+d}+\cdots+\frac1{a+nd}$$

is not an integer. (Display (1), p. 1; the German summary on p. 8 states the same: "Es seien $a,d,n$ beliebige positive ganze Zahlen, dann ist $\frac1a+\frac1{a+d}+\cdots+\frac1{a+nd}$ keine ganze Zahl.")

The introduction (p. 1) records the earlier theorems being generalized: Theisinger proved that the partial sums $\sum_{m=2}^n1/m$ of the harmonic series are never integers (Monatshefte für Math. u. Phys. 26 (1915), 135); Obláth proved that $\sum_{m=k}^na_m/m$ is not an integer when the $a_m$ are positive integers with $(a_m,m)=1$ (Mat. Fiz. Lapok 27 (1918), 93; the German summary, p. 8, has the lower limit $m=2$); Kürschák gave an elementary proof that $\sum_{m=k}^n1/m$ is never an integer, for whatever positive integers $k$ and $n$ (the print states no restriction; read literally, the claim fails at $k=n=1$, where the sum is $1$) (Mat. Fiz. Lapok 27 (1918), 299). The case $d=1$ of the theorem is Kürschák's theorem.

Proof pointer and sketch (pp. 1--7)

• Reduction (p. 1): one may assume $(a,d)=1$, since factoring out the reciprocal of the greatest common divisor leaves a sum of the same shape with coprime parameters.
• Lemma (2) (p. 2), for $d\ge4$: among $a+d,\,a+2d,\ldots,a+nd$ some term is divisible by a prime power $p^\alpha>n$. Given the lemma, if $a+kd$ is divisible by $p^\alpha>n$ then no other $a+k'd$ with $|k-k'|<n$ is (otherwise $p^\alpha\mid(k-k')d$ with $(d,p)=1$), and $p^\alpha\nmid a$ (otherwise $p^\alpha\mid kd$ with $k<p^\alpha$); writing $!(a+nd)=(a+d)(a+2d)\cdots(a+nd)$ (display (3)) and putting the sum over the common denominator $a\cdot!(a+nd)$ (display (4)), the numerator term $a\cdot!(a+nd)/(a+kd)$ is divisible by a lower power of $p$ than every other numerator term and than the denominator, so the quotient is not an integer.
• Proof of the lemma (pp. 2--6): suppose every term is divisible only by prime powers $p^\alpha\le n$; then $!(a+nd)/n!$ (display (5)) is bounded above by $\prod_{p\le n}p\cdot\prod_{p\le\sqrt n}p\cdots$ (display (6)), while $!(a+nd)/n!=\prod_{i=1}^n(a/i+d)>d^n\ge4^n$ for $d\ge4$ (display (7)), giving (8). The product of primes is bounded through the prime factorization of binomial coefficients $\binom{2n}{n}$ (displays (9)--(13)), using that $\binom{2n}{n}<4^{n-1}$ for $n\ge5$ and an induction on the sequence $a_k=\lceil n/2^k\rceil$; the resulting inequality contradicts (8). The cases $n\le10$ are checked by computation.
• Special cases (pp. 6--7): $d=1$ is Kürschák's theorem; $d$ odd (in particular $d=3$) follows as in Kürschák's proof from the highest power of $2$ dividing a term, which divides exactly one term of the progression; $d=2$ uses the highest power of $3$ in the same way: since $1/a$ exceeds every other term, an integer sum needs $n\ge a$, so the last term is at least $3a$ and every odd number from $a$ to $3a$ occurs ($a$ being odd), among them a power of $3$. A closing remark (p. 7), without proof, states that similar but somewhat longer computations prove, in these special cases too, that some term $a+kd$ of $a,a+d,\ldots,a+nd$ contains some prime $p$ to a higher power than every other term; that the lemma also holds in these cases (except $d=1$); and that Obláth's theorem generalizes to $\sum a_k/(a+kd)\notin\mathbb Z$ when $(a_k,a+kd)=1$, provable by his method.

These steps were read for structure on the page images and are recorded as a sketch; no complete rewritten proof and no independent review exist here.

Relation to Problem 287

If a representation $1=\sum_{i=1}^k1/n_i$ by distinct integers $1<n_1<\cdots<n_k$ had every consecutive gap equal to $1$, its denominators would form a progression $a,a+1,\ldots,a+(k-1)$ and the theorem with $d=1$ (Kürschák's case) would be contradicted; so every such representation has $\max(n_{i+1}-n_i)\ge2$. This is the "lower bound of $\ge2$" in the site's commentary. The theorem concerns complete arithmetic progressions only: a set of denominators whose gaps mix $1$ and $2$ is not a progression, so beyond excluding all gaps $1$ ($d=1$) and all gaps $2$ ($d=2$), the theorem says nothing about the gap-$3$ question itself.

Bears on. #287 (the gap-$\ge2$ fact through the case $d=1$; not the gap-$\ge3$ statement); #288 (through the case $d=1$, no interval of two or more consecutive integers has an integer reciprocal sum; nothing about sums over two intervals beyond two adjacent ones, whose union is a single interval).