---

Source. Section 1, printed p. 93 (the claim for every $k$ and the announcement that only $k=1$ is proved); section 2, pp. 94--96 (the proof for $k=1$). Read on the page images (physical PDF pp. 1--4).

Statement

Let $p_n$ be the $n$-th prime. The paper claims that for every $k=1,2,3,\ldots$ the sum of the series

$$\sum_{n=1}^{\infty}\frac{p_n^k}{n!}$$

(series (1), p. 93) is irrational, and proves:

Theorem ($k=1$). $\displaystyle\sum_{n=1}^{\infty}\frac{p_n}{n!}$ is irrational.

For $k\ge2$ the paper gives no proof: "la démonstration étant assez compliquée pour $k>1$, je ne donnerai au § 2 que la démonstration pour $k=1$" (p. 93). The cases $k\ge2$ were first proved in print by Schlage-Puchta 2007, Theorem 3, in the stronger form that $1,S_0,S_1,S_2,\ldots$ are $\mathbb{Q}$-linearly independent, $S_k=\sum p_n^k/n!$. This page attributes nothing beyond $k=1$ to the 1958 paper.

Premises

• (P1) $p_n=o(n^2)$. The paper uses it as "puisque $p_k=o(k^2)$" (p. 95). It follows from $p_n\sim n\log n$, the prime number theorem, which the paper uses on p. 96; Chebyshev's elementary bound $p_n\ll n\log n$ also gives it.
• (P2) The fractional parts $\{p_n/n\}$, $n\ge1$, are dense in $(0,1)$. This is statement (2) of p. 94, proved on its own page from the prime number theorem with remainder (Landau) and the Pólya–Szegő density criterion.

Complete rewritten proof of the case $k=1$

Write $[x]$ for the integer part and $\{x\}=x-[x]$.

Step 1 (an integer tail). Suppose $\sum_{n\ge1}p_n/n!=a/b$ with positive integers $a,b$. Fix an integer $k>b$. Then $b$ divides $(k-1)!$, so $(k-1)!\,a/b$ is an integer; and $(k-1)!\sum_{n\le k-1}p_n/n!$ is an integer because $(k-1)!/n!$ is an integer for every $n\le k-1$. Their difference

$$T_k:=(k-1)!\sum_{n\ge k}\frac{p_n}{n!} =\frac{p_k}{k}+\frac{p_{k+1}}{k(k+1)}+\frac{p_{k+2}}{k(k+1)(k+2)}+\cdots$$

is therefore an integer, and it is positive because every term is positive; hence $T_k\ge1$ for every $k>b$. (Paper, p. 94: the display "$\cdots=(k-1)!\,a/b$ est un entier positif"; as printed, the equality omits the integer $(k-1)!\sum_{n\le k-1}p_n/n!$ subtracted above.)

Step 2 (the tail after the first term tends to zero). Write $T_k=p_k/k+R_k$ with

$$R_k:=\sum_{j\ge1}\frac{p_{k+j}}{k(k+1)\cdots(k+j)}.$$

Put $\varepsilon_k:=\sup_{m\ge k}p_m/m^2$; by (P1), $\varepsilon_k\to0$, and $p_{k+j}\le\varepsilon_k(k+j)^2$ for all $j\ge1$. For $j=1$ the factor $(k+1)^2/(k(k+1))=(k+1)/k$ is at most $2$. For $j\ge2$ and $k\ge2$ we have $k(k+1)\cdots(k+j)\ge k^{j+1}$ and $k+j\le kj$, so

$$\frac{(k+j)^2}{k(k+1)\cdots(k+j)}\le\frac{j^2}{k^{j-1}}\le\frac{j^2}{2^{j-1}}, \qquad\sum_{j\ge2}\frac{j^2}{2^{j-1}}=11 .$$

Hence $0<R_k\le13\,\varepsilon_k\to0$. (Paper, p. 95: "cette inégalité ne peut avoir lieu pour $k$ suffisamment grand puisque $p_k=o(k^2)$"; the explicit bound is supplied here.)

Step 3 (a lower bound for the fractional part). By Step 1, $\{p_k/k\}+R_k=T_k-[p_k/k]$ is an integer; it is positive because $R_k>0$; so

$$\left\{\frac{p_k}{k}\right\}+R_k\ge1\qquad\text{for every }k>b .$$

(Paper, p. 94: "$p_k/k-[p_k/k]+p_{k+1}/(k(k+1))+\cdots\ge1$".)

Step 4 (contradiction). By (P2) there are infinitely many $k$ with $\{p_k/k\}\le1/2$ (paper, p. 94: "il existe une infinité de $k$ tels que $p_k/k-[p_k/k]\le1/2$"). For every such $k>b$, Step 3 gives $R_k\ge1/2$, which contradicts Step 2 once $k$ is large. Hence $\sum p_n/n!$ is irrational. $\blacksquare$

Remarks

• Primality enters only through (P1) and (P2). The paper says so indirectly on p. 96 ("contrairement au cas traité au § 2, la démonstration de ce théorème [section 3] utilise pleinement le fait que les $p_n$ sont premiers") and isolates the argument as the proposition on p. 95.
• For $k\ge2$ the growth premise fails for $c_n=p_n^k$ ($p_n^k/n^2\to\infty$), so this argument does not extend; the paper offers no other argument.
• The only non-elementary input is the remainder term behind (P2); the paper (p. 95) asks whether a more elementary proof of the density can be found.

Verification

This full reconstruction is author-recorded. It contains every deduction of section 2 (pp. 94--96); Step 2 expands the paper's one-line estimate, and statement (2) is proved on its own page with the prime number theorem with remainder as an unread external premise (its statement compared with the paper's citation of Landau; Landau's text not consulted and no proof inspected) and the Pólya–Szegő criterion proved there. No independent review has been filed; until a whole-claim review of this page and the density page is filed under this card's evidence/verify/, the proof is not independently accepted compilation proof coverage.

Bears on. #251 (context: the site's remark on problem 251 attributes $\sum p_n^k/n!$ for every $k\ge1$ to this paper; the paper proves $k=1$).