Infinite tail representation with controlled reciprocal mass

---

Source and authorship. Fredy Yip, On a problem of Erdős and Ingham, arXiv:2512.16528v1, 18 December 2025, printed/physical p. 2, immediately after Theorem 1.3 (arXiv v1 PDF). The source asserts the simultaneous infinite-set, arbitrary-tail, and mass refinement. The proof below supplies its omitted scheduling argument from the source's Lemma 2.1. It is a project-authored completion of that existing assertion. It was independently reviewed together with the complete natural-language chain; the refinement audit and full-proof review retain that report. This is not a claim that these details are printed in v1 or that the preprint is accepted.

Exact statement

For every real $t\ne0$, every $\lambda\in\mathbb C$, every positive integer $N$, and every real $\delta>0$, there is an infinite set $S\subseteq\mathbb Z_{\ge\max\{2,N\}}$ such that

$$\sum_{n\in S}\frac1n\le |\lambda|+\delta, \qquad \sum_{n\in S}n^{-(1+it)}=\lambda. \tag{R1}$$

The complex series is absolutely convergent. The lower bound $\max\{2,N\}$ retains both the underlying Theorem 1.3 restriction $S\subseteq\mathbb Z_{\ge2}$ and the arbitrary cutoff in the following sentence. The set may depend on all four parameters. No assertion is made for $t=0$.

Source-derived finite-block interface

Fix $t\ne0$ and put $K=1+|1+it|$. Lemma 2.1 (p. 2) gives, for every positive integer $B$ and every $c\in\mathbb C$, a finite set $A\subseteq\mathbb Z_{\ge B}$ satisfying

$$\left|c-\sum_{n\in A}n^{-(1+it)}\right|\le K|c|^2, \qquad \sum_{n\in A}\frac1n\le |c|. \tag{R2}$$

We use the construction in its proof, which makes $A$ nonempty whenever $c\ne0$. To check this extra interface fact, write $c=|c|e^{i\theta}$. The equation $-t\log x=\theta+2\pi j$ has arbitrarily large positive real solutions $x$ as integers $j$ tend in the appropriate direction. Choose one with $x\ge\max\{B,|c|^{-1},|c|^{-2}\}$. Then $s=\lfloor x|c|\rfloor\ge1$ and $A=[x,x+s)\cap\mathbb Z$ has exactly $s$ elements. For example, they are $\lceil x\rceil,\ldots,\lceil x\rceil+s-1$, whether or not $x$ is an integer. All are at least $B$. The lemma's proof shows that this block satisfies (R2), with this same $K$, for every such cutoff. For $c=0$ the lemma instead permits $A=\varnothing$; the nonzero branch below never invokes that case.

Proof when the target is nonzero

Suppose $\lambda\ne0$. Write $L=|\lambda|>0$ and $M=\max\{2,N\}$. Choose

$$q=\frac{\delta}{2(L+\delta)}, \qquad \rho=\frac qK, \qquad \alpha=1-q. \tag{R3}$$

These choices give $0<q<1/2$, $\rho>0$, $K\rho=q<1$, and

$$\frac L\alpha =L+\frac{L\delta}{2L+\delta} <L+\delta. \tag{R4}$$

We construct a nonempty finite block $A_k$ at every positive integer stage $k$. Before the first stage put $w_1=\lambda$. Inductively, assuming $w_k\ne0$, define

$$v_k=|w_k|, \qquad a_k=\min\{\rho,v_k/2\}, \qquad c_k=a_k\frac{w_k}{v_k}. \tag{R5}$$

Thus $a_k>0$, $c_k\ne0$, and $c_k$ points in the direction of $w_k$. At the first stage take $B_1=M$. At each later stage take an integer $B_k\ge M$ greater than every element of $A_1\cup\cdots\cup A_{k-1}$. This is possible since that union is finite. Use the nonempty construction in (R2) with cutoff $B_k$ and target $c_k$; call its block $A_k$. Define

$$z_k=\sum_{n\in A_k}n^{-(1+it)}, \qquad e_k=c_k-z_k, \qquad w_{k+1}=w_k-z_k. \tag{R6}$$

The blocks are disjoint and strictly ordered. Equations (R2), (R3), and $a_k\le\rho$ give

$$|e_k|\le K a_k^2\le q a_k, \qquad \sum_{n\in A_k}\frac1n\le a_k. \tag{R7}$$

Since $a_k\le v_k/2$, radial alignment gives $|w_k-c_k|=v_k-a_k$. Both triangle inequalities applied to $w_{k+1}=(w_k-c_k)+e_k$ therefore give

$$\begin{aligned} v_{k+1} &\ge v_k-a_k-|e_k| \ge v_k-(1+q)a_k \ge \frac{\alpha v_k}{2}>0, \\[2pt] v_{k+1} &\le v_k-a_k+|e_k| \le v_k-\alpha a_k. \end{aligned} \tag{R8}$$

The strict lower bound proves inductively that every $w_k$ is nonzero. Consequently every $a_k$ is positive and every block is nonempty; the construction cannot terminate. The upper bound shows that the moduli strictly decrease. While $v_k>2\rho$, it gives the fixed decrement $v_{k+1}\le v_k-\alpha\rho$. That phase must end after finitely many steps, since otherwise iterating this inequality would make a modulus negative. Once $v_k\le2\rho$, we have $a_k=v_k/2$, and (R8) gives

$$v_{k+1}\le\frac{1+q}{2}v_k. \tag{R9}$$

The multiplier lies strictly between $1/2$ and $3/4$. Thus the small-modulus phase persists, $v_k\to0$, and $w_k\to0$.

To control the total mass, telescope the upper bound in (R8). For every positive integer $J$,

$$\alpha\sum_{k=1}^{J}a_k \le\sum_{k=1}^{J}(v_k-v_{k+1}) =L-v_{J+1}\le L. \tag{R10}$$

The partial sums on the left increase and are bounded. Passing to their limit and using (R4) yields

$$\sum_{k\ge1}a_k\le\frac L\alpha<L+\delta. \tag{R11}$$

Set $S=\bigcup_{k\ge1}A_k$. It lies in $\mathbb Z_{\ge M}$ and is infinite: its first $J$ pairwise disjoint nonempty blocks already contain at least $J$ distinct integers. By disjointness, nonnegativity, and (R7), (R11),

$$\sum_{n\in S}\frac1n =\sum_{k\ge1}\sum_{n\in A_k}\frac1n \le\sum_{k\ge1}a_k \le\frac L\alpha<L+\delta. \tag{R12}$$

Here the equality of nonnegative sums follows by taking suprema of finite sums: every finite subset of $S$ is contained in finitely many blocks, and each finite collection of blocks has finite union. Since $|n^{-(1+it)}|=1/n$, (R12) proves absolute convergence of the complex series. The recurrence (R6) gives

$$\sum_{n\in A_1\cup\cdots\cup A_J}n^{-(1+it)} =\lambda-w_{J+1}\longrightarrow\lambda. \tag{R13}$$

The ordered blocks exhaust $S$, and the absolute mass of the remaining blocks tends to zero by (R11). Therefore (R13) identifies the full complex sum with $\lambda$. This proves every requirement in (R1) for a nonzero target.

Proof when the target is zero

Now suppose $\lambda=0$. Choose an integer $m\ge M$ large enough that $2/m<\delta$, and put

$$\beta=-m^{-(1+it)}, \qquad \varepsilon=\delta-2/m>0. \tag{R14}$$

The target $\beta$ is nonzero and has modulus $1/m$. Apply the already proved nonzero case with the same $t$, cutoff $m+1$, target $\beta$, and slack $\varepsilon$. It supplies an infinite set $T\subseteq\mathbb Z_{\ge m+1}$ with

$$\sum_{n\in T}n^{-(1+it)}=\beta, \qquad \sum_{n\in T}\frac1n\le\frac1m+\varepsilon. \tag{R15}$$

Then $S=\{m\}\cup T$ is infinite, the union is disjoint, and $S\subseteq\mathbb Z_{\ge M}$. Absolute convergence permits adjoining the singleton, so

$$\sum_{n\in S}n^{-(1+it)}=m^{-(1+it)}+\beta=0, \qquad \sum_{n\in S}\frac1n\le\frac2m+\varepsilon=\delta. \tag{R16}$$

This proves (R1) also for $\lambda=0$. $\square$

Exact transfer to the infinite-sequence question

For Problem 967, take, for example, $t=1$, $\lambda=-1$, $N=2$, and $\delta=1$ in the nonzero branch. The resulting infinite $S\subseteq\mathbb Z_{\ge2}$ has finite reciprocal mass and complex sum $-1$. Enumerate $S$ in increasing order by repeatedly choosing its least unused member. Every finite stage leaves a member because $S$ is infinite. Every $n\in S$ eventually appears because there are only finitely many positive integers at most $n$. Thus this is an enumeration $1<a_1<a_2<\cdots$ of all of $S$. Absolute convergence, proved in (R12), gives

$$\sum_{k\ge1}\frac1{a_k}<\infty, \qquad 1+\sum_{k\ge1}\frac1{a_k^{1+i}}=0. \tag{R17}$$

One admissible sequence and one real value of $t$ refute the universal nonvanishing assertion. The zero-target branch is unnecessary for this specialization; it completes the full source assertion for arbitrary $\lambda$.

Scope and review record

The asserted refinement is Yip's; (R3)--(R16) are the separately authored schedule, estimates, and zero-target reduction supplied here. The proof uses only Lemma 2.1, the construction in its proof, and the elementary limit arguments written above. It does not rely on the printed theorem's two misprints or on Erdős--Ingham's contextual Tauberian theorem.

The finite-set Conjecture 3.1 and fixed-set Question 3.2 for $S=\{2,3,5\}$ remain separate and are left open in v1. This construction always supplies an infinite set and gives no solution of those finite questions. No Lean code was run or repaired. The complete natural-language chain, including this separately attributed refinement, passed independent strong mathematical and source review, retained as the full-proof review. That review gives no formal, peer-review-acceptance, recursive external-proof, or canonical-integration credit.

Bears on. Problem 967: the transfer above gives, for $t=1$, an infinite sequence of integers $1<a_1<a_2<\cdots$ with $\sum_k a_k^{-1}<\infty$ and $1+\sum_k a_k^{-(1+i)}=0$, so the problem's assertion fails for that $t$. The finite case is untouched.