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Source. Theorem 3, p. 2 of the arXiv PDF; proof pp. 7--9, using Lemmas 3--6 (pp. 5--6) and the quoted Lemmas 1--2 (p. 2). Read in the text layer and checked on the rendered pages.

Statement

For an integer k≥0k\ge0 let

Sk:=∑n=1∞pnkn!,S_k:=\sum_{n=1}^{\infty}\frac{p_n^k}{n!},

pnp_n the nn-th prime. Then the real numbers 1,S0,S1,S2,…1,S_0,S_1,S_2,\ldots are Q\mathbb{Q}-linearly independent.

In particular every SkS_k with k≥1k\ge1 is irrational (S0=e−1S_0=e-1). The paper's own framing (p. 2): Erdős stated in 1958 that SkS_k is irrational and proved k=1k=1; "it appears that, for k>1k>1, no proof has appeared in print". The statement for all k≥1k\ge1 therefore has two sources: k=1k=1 in Erdős 1958 and k≥2k\ge2 here.

Reduction (p. 7)

A nontrivial relation c+∑k≤KakSk=0c+\sum_{k\le K}a_kS_k=0 with rational coefficients has some ak≠0a_k\ne0, and then S:=∑ν≥1P(pν)/ν!S:=\sum_{\nu\ge1}P(p_\nu)/\nu! with P=∑akxk≠0P=\sum a_kx^k\ne0 is rational; clearing denominators, the theorem reduces to the irrationality of ∑ν≥1P(pν)/ν!\sum_{\nu\ge1}P(p_\nu)/\nu! for every nonzero P∈Z[x]P\in\mathbb{Z}[x]. The paper states this reduction (p. 7) as "It suffices to show that S=∑ν=1∞P(pν)/ν!S=\sum_{\nu=1}^{\infty}P(p_\nu)/\nu! is irrational for every polynomial PP with integral coefficients which does not vanish identically."

Dependencies

  • Lemma 1 (Weyl–van der Corput) and Lemma 2 (Erdős–Turán), p. 2, quoted from the literature ([3, Theorem 2.8], [6, II, Theorem 2.5]): an exponential-sum bound for functions with controlled (q+2)(q+2)-nd derivative, and the discrepancy bound DN≪N/H+∑1≤h≤H∣∑n≤Ne(hxn)∣D_N\ll N/H+\sum_{1\le h\le H}\big|\sum_{n\le N}e(hx_n)\big|.
  • Lemma 3 (p. 5), "consequence of Selberg's sieve, confer, e.g. [4, Theorem 5.1]" (Halberstam–Richert): if 0≤a1<⋯<ak<N0\le a_1<\cdots<a_k<N are integers and N⊆[x,2x]\mathcal{N}\subseteq[x,2x] is a set of integers nn with every n+ain+a_i prime, then ∣N∣≤ckxlog⁡k+1x∏p(1+1p)k+2−ν(p)|\mathcal{N}|\le\frac{c_kx}{\log^{k+1}x}\prod_p(1+\frac1p)^{k+2-\nu(p)}, ν(p)\nu(p) the number of distinct residues modulo pp among the shifts; in particular ∣N∣≪kxlog⁡2k+2x/log⁡k+1x|\mathcal{N}|\ll_kx\log_2^{k+2}x/\log^{k+1}x, log⁡2\log_2 the iterated logarithm. (The statement writes the shifts as aia_i and the residue set as {0,Δ0,Δ0+Δ1,…}\{0,\Delta_0,\Delta_0+\Delta_1,\ldots\}; the range of the product over pp is not specified in the statement, and the "in particular" bound is printed without the argument of log⁡2\log_2.)
  • Lemma 4 (p. 5): a nonzero F∈Z[x0,…,xk]F\in\mathbb{Z}[x_0,\ldots,x_k] has F(δn,…,δn+k)≠0F(\delta_n,\ldots,\delta_{n+k})\ne0 for almost all nn, where δn=pn+1−pn\delta_n=p_{n+1}-p_n.
  • Lemma 5 (p. 6): if P,Q∈k[X1,…,Xn]P,Q\in k[X_1,\ldots,X_n] over a field kk, ν≠0\nu\ne0 is an integer, and the polynomial νX1P+P Q+−P+Q\nu X_1P+P\,Q^{+}-P^{+}Q vanishes identically, where P+=P(X2,…,Xn+1)P^{+}=P(X_2,\ldots,X_{n+1}) and Q+=Q(X2,…,Xn+1)Q^{+}=Q(X_2,\ldots,X_{n+1}) are the index shifts, then PP vanishes identically. Proof on p. 6 by setting X1=0X_1=0 and eliminating variables.
  • Lemma 6 (p. 6; proof p. 7): for a nonconstant Q∈Z[X]Q\in\mathbb{Z}[X] of degree dd with coefficients bounded by MM, the discrepancy DD of the sequence Q(pn/n) mod 1Q(p_n/n)\bmod1, x≤n≤2xx\le n\le2x, satisfies D≪xe−clog⁡x+M1/3x2/3log⁡d/3xD\ll xe^{-c\sqrt{\log x}}+M^{1/3}x^{2/3}\log^{d/3}x. The proof replaces pnp_n by the inverse function of li\mathrm{li} at nn using the prime number theorem with the classical error term, then applies Lemma 2 and the case q=0q=0 of Lemma 1.

Proof structure (pp. 7--9)

Step 1. Assume S=∑ν≥1P(pν)/ν!S=\sum_{\nu\ge1}P(p_\nu)/\nu! is rational, deg⁡P=k\deg P=k. For large nn, n!Sn!S is an integer, so the scaled tail ∑ν>nP(pν)/((n+1)⋯ν)\sum_{\nu>n}P(p_\nu)/((n+1)\cdots\nu) is an integer. Since pν∼νlog⁡νp_\nu\sim\nu\log\nu, only the first few terms matter: the paper writes, with ∥⋅∥\|\cdot\| the distance to the nearest integer,

∥∑ν=1k−1P(pn+ν)(n+1)⋯(n+ν)∥≪log⁡knn\left\|\sum_{\nu=1}^{k-1}\frac{P(p_{n+\nu})}{(n+1)\cdots(n+\nu)}\right\| \ll\frac{\log^kn}{n}

for large nn, and calls the truncated sum F(0)(n)F^{(0)}(n).

Step 2 (p. 8). Writing pn+i=pn+δn+⋯+δn+i−1p_{n+i}=p_n+\delta_n+\cdots+\delta_{n+i-1} and expanding 1/((n+1)⋯(n+i))1/((n+1)\cdots(n+i)) in powers of 1/n1/n gives

F(0)(n)=∑ν=1k∑μ=1νPνμ(0)(δn,…,δn+k−1) pnνnμ+R(n),F^{(0)}(n)=\sum_{\nu=1}^{k}\sum_{\mu=1}^{\nu} P^{(0)}_{\nu\mu}(\delta_n,\ldots,\delta_{n+k-1})\,\frac{p_n^\nu}{n^\mu}+R(n),

where R(n)R(n) denotes any error ≪log⁡cn/n\ll\log^cn/n for almost all nn (so that δnR(n)=R(n)\delta_nR(n)=R(n)).

Step 3. Pairs (ν,μ)(\nu,\mu) are ordered by ν−μ\nu-\mu, the growth exponent of pnν/nμp_n^\nu/n^\mu, then by ν\nu. With (ν0,μ0)(\nu_0,\mu_0) the maximal pair present in F(i)F^{(i)}, the recursion

F(i+1)(n)=Pν0μ0(i)(δn+1,…,δn+k)F(i)(n)−Pν0μ0(i)(δn,…,δn+k−1)F(i)(n+1)F^{(i+1)}(n)=P^{(i)}_{\nu_0\mu_0}(\delta_{n+1},\ldots,\delta_{n+k})F^{(i)}(n) -P^{(i)}_{\nu_0\mu_0}(\delta_n,\ldots,\delta_{n+k-1})F^{(i)}(n+1)

keeps ∥F(i)(n)∥=R(n)\|F^{(i)}(n)\|=R(n) and removes the leading monomial; Lemma 5 shows that the new coefficient of pnν0−1/nμ0p_n^{\nu_0-1}/n^{\mu_0} does not vanish identically. After finitely many steps only pairs with μ=ν\mu=\nu remain, at least one with a nonzero coefficient.

Step 4. Clearing denominators, there are ℓ\ell and polynomials Qi∈Z[X1,…,Xℓ]Q_i\in\mathbb{Z}[X_1,\ldots,X_\ell] with Qℓ≠0Q_\ell\ne0 and

∥∑i=1ℓQi(δn,…,δn+ℓ) pnini∥=R(n).\left\|\sum_{i=1}^{\ell}Q_i(\delta_n,\ldots,\delta_{n+\ell})\, \frac{p_n^i}{n^i}\right\|=R(n).

By Lemma 4 some Qi(δn,…,δn+ℓ)≠0Q_i(\delta_n,\ldots,\delta_{n+\ell})\ne0 for almost all nn, and for almost all nn no δn+j\delta_{n+j} exceeds log⁡2n\log^2n; so (p. 9, formula (2)) for almost all nn there are integers a1,…,aℓa_1,\ldots,a_\ell, not all zero, with 0<∣ai∣<log⁡An0<|a_i|<\log^An, such that ∥∑iai(pn/n)i∥≪e−clog⁡n\big\|\sum_ia_i(p_n/n)^i\big\|\ll e^{-c\sqrt{\log n}}.

Step 5. Pigeonhole: one tuple (ai)(a_i) serves at least x/log⁡ℓAxx/\log^{\ell A}x integers n≤xn\le x, and the paper ends (p. 9): "This clearly contradicts Lemma 6". The count is not written out; it would run through Lemma 6 with M=log⁡AxM=\log^Ax, which bounds the number of n∈[x,2x]n\in[x,2x] with ∥Q(pn/n)∥≤e−clog⁡x\|Q(p_n/n)\|\le e^{-c\sqrt{\log x}} by o(x/log⁡ℓAx)o(x/\log^{\ell A}x). ■\blacksquare

Where scrutiny would begin

Recorded for a future review; none has been made. (i) The truncation in step 1 is printed with upper index k−1k-1 (k=deg⁡Pk=\deg P), but the term of index ν\nu has size of order log⁡kn/nν−k\log^kn/n^{\nu-k}, so the term of index kk is of order log⁡kn\log^kn, larger than the stated error log⁡kn/n\log^kn/n, and the truncation must run at least to kk. (In step 2 the outer index ν\nu is the power of pnp_n, not the truncation index.) (ii) The bookkeeping of "almost all nn" through the finitely many recursion steps and the size of the truncation error R(n)R(n), including its interaction with the gap bound δn+j≤log⁡2n\delta_{n+j}\le\log^2n. (iii) The pigeonhole count against Lemma 6: the constants AA and ℓ\ell depend on PP, and M=log⁡AxM=\log^Ax must keep M1/3x2/3log⁡ℓ/3x=o(x/log⁡ℓAx)M^{1/3}x^{2/3}\log^{\ell/3}x=o(x/\log^{\ell A}x), which it does. (iv) In Lemma 3 the constant ckc_k and the range of the product are not made explicit; Lemma 4 uses only the "in particular" bound.

Bears on. #251 (context: this is the k≥2k\ge2 half of the theorem the site's remark attributes wholly to Erdős 1958; it says nothing about ∑pn/2n\sum p_n/2^n).