Reciprocal Mersenne Subseries

A finite estimate on common observation scales

A divisor majorant becomes a sum of geometric tails, one for each divisor. The positive offsets $r$ with $d\mid N+r$ are $d-(Nmodd),2d-(Nmodd),\dots$, and hence

The cover argument raises a base-two tail to a fractional power $\alpha$, replacing its geometric factor by $B=2^{\alpha }$. The estimate must therefore allow $1<B\le 2$. It must also keep the modulus, first dyadic scale and number of scales separate. In the mixed theorem, the weighted construction chooses these parameters after the finite divisibility requirements have been fixed.

For $1<B\le 2$, positive integers $L,d,M$, and an integer $R\ge 0$, put

To read this as sampling, first choose $j$ uniformly from $R,\dots ,R+M-1$. Conditional on that scale, choose $m$ uniformly from $1,\dots ,2^j$ and observe $N=Lm$. The scales have equal total weight; the observations within a longer scale have smaller individual weights. Repeated occurrences of an integer $N$ retain their contributions to the sum. This is the distribution used for both parts of the mixed proof.

The finite estimate

equivalently bounds $(B-1){\mathcal{D}}_{L;R,M}{w}_{B,d}$ by $(1+4L/M)/d$. This normalised bound is uniform as $B\downarrow 1$. The unnormalised right side grows like $(B-1{)}^{-1}$. In the cover argument $B=2^{\alpha }$ can approach $1$, so the factor $(2^{\alpha }-1{)}^{-1}$ must remain in the summability condition. The ratio $L/M$ measures the cost of incomplete modular periods.

Fix $T=2^j$. We compare $d$ with the largest sampled multiple $LT$. For $d\le LT$ we use a complete-orbit bound with a remainder term; for $d>2LT$ no residue is reduced modulo $d$. For fixed $d$, doubling $T$ moves from $d>2LT$ to $d\le LT$ after at most one intermediate scale; hence $LT<d\le 2LT$ occurs for at most one dyadic length. When $d\le LT$, the complete orbit has length $d/g$, with $g=(L,d)$, and total weight $1/(B^g-1)$. Splitting the sample into complete cycles—possibly none—and one remaining piece gives

Across dyadic lengths satisfying $d\le L{2}^j$, the reciprocal-length errors sum to at most $2L/[d(B-1)]$.

When $d>2LT$, we have $Lmmodd=Lm$ and $2Lm\le d-1$; hence $\sum _{i=0}^{d-1}{B}^i\ge d{B}^{(d-1)/2}\ge d{B}^{Lm}$. Each term $w_{B,d}(Lm)$ is then at most $1/[d(B-1)]$.

Finally, at most one dyadic length satisfies $LT<d\le 2LT$. For that length the geometric sum and convexity give

The first inequality follows from convexity of $x\mapsto B^x-1$, which vanishes at $0$: the ratio is at most $LT/d$. The last uses $(B^L-1)/(B-1)=\sum _{i=0}^{L-1}{B}^i\ge B^{L-1}$ and $B\le 2$. Summing the main terms and the two error bounds proves (13). The argument displayed here is an ordinary proof; the estimate is also proved in Lean. Nonnegative interchange permits summation against any coefficients $c_d$ with $\sum _d{c}_d/d<\mathrm{\infty }$.