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Write $A(x)=\sum_{n\ge1}\varphi(n)x^n$, $S=A(1/2)$, and $B(x)=x/(1-x)^2$. These are the conventions of the parent sketch. All new arguments here are author-recorded and unreviewed. The obstruction concerns the literal rational sieve sums defined below, with their reduced denominators.

The exact family and its entire remainder

For a squarefree positive integer $Q$ define

$$M_Q(x)=\sum_{d\mid Q}\mu(d)B(x^d).$$

Then, for $|x|<1$,

$$M_Q(x)=\sum_{\substack{n\ge1\\(n,Q)=1}}A(x^n). \tag{1}$$

Indeed, the coefficient of $x^m$ on the left is $m\sum_{d\mid(m,Q)}\mu(d)/d$. Write $m=m_0m_1$, where every prime factor of $m_0$ divides $Q$ and $(m_1,Q)=1$. The coefficient on the right is $\sum_{n\mid m_1}\varphi(m/n) =\varphi(m_0)\sum_{n\mid m_1}\varphi(m_1/n) =\varphi(m_0)m_1$, the same quantity. Absolute convergence follows from $\varphi(n)\le n$ and justifies the analytic identity.

From now on let $Q$ be squarefree with largest prime factor $P\ge5$, and let $q$ be the least prime not dividing $Q$. The prime divisors of $Q$ need not be consecutive. Put

$$M_Q=M_Q(1/2) =\sum_{d\mid Q}\frac{\mu(d)2^d}{(2^d-1)^2}, \qquad R_Q=M_Q-S.$$

The entire omitted series is positive and obeys

$$2^{-q}\le R_Q =\sum_{\substack{n>1\\(n,Q)=1}} A(2^{-n}) \le\frac{2^{1-q}}{(1-2^{-q})^2}. \tag{2}$$

The least admissible index is $q$, and $A(t)\ge t$ for $0<t<1$. For the upper bound use $A(t)\le t/(1-t)^2$ and include all $n\ge q$. Both bounds concern the full infinite remainder in the same parameter $Q$ as the rational approximant.

Which denominator factors cannot cancel

Let $\Psi_Q(X)$ denote the $Q$th cyclotomic polynomial and put $C_Q=\Psi_Q(2)$. Define

$$G_Q=\frac{C_Q}{\gcd(C_Q,P)}.$$

Lemma 1. Every prime factor $\ell$ of $G_Q$ satisfies $\operatorname{ord}_{\ell}(2)=Q$. Moreover

$$G_Q\ge\frac{C_Q}{P}>\frac{2^{\varphi(Q)-2}}P. \tag{3}$$

Proof. The Möbius formula for the cyclotomic value is

$$C_Q=\prod_{d\mid Q}(2^d-1)^{\mu(Q/d)}. \tag{4}$$

It is an odd positive integer. Fix an odd prime $\ell$ and put $m=\operatorname{ord}_{\ell}(2)$. If $m\nmid Q$, no factor in (4) has nonzero $\ell$-adic valuation. If $m\mid Q$, write $Q=mh$. The squarefreeness of $Q$ makes $(m,h)=1$, and $\ell\nmid m$ since $m\mid\ell-1$. The elementary valuation formula

$$v_\ell(2^{me}-1)=v_\ell(2^m-1)+v_\ell(e) \qquad(e\ge1) \tag{5}$$

gives

$$v_\ell(C_Q)= \begin{cases} v_\ell(2^Q-1),&m=Q,\\ 1,&h=\ell,\\ 0,&\text{otherwise}. \end{cases} \tag{6}$$

For completeness, substitute $d=me$ in (4). The constant term in (5) sums to zero over $e\mid h$ when $h>1$. Because $h$ is squarefree, the remaining sum is $\sum_{e\mid h,\,\ell\mid e}\mu(h/e)$, which equals one exactly for $h=\ell$. Formula (5) follows by the binomial expansion of $(1+\ell^a u)^\ell$ for odd $\ell$ and $\ell\nmid u$, followed by induction and the case of an exponent coprime to $\ell$.

In the nonprimitive case $Q=m\ell$ with $m\mid\ell-1$, every other prime divisor of $Q$ is smaller than $\ell$. Thus $\ell=P$, and its exponent in $C_Q$ is exactly one. Removing $\gcd(C_Q,P)$ therefore removes every possible nonprimitive factor.

Finally (4) and $\sum_{d\mid Q}\mu(Q/d)d=\varphi(Q)$ imply

$$C_Q=2^{\varphi(Q)} \prod_{d\mid Q}(1-2^{-d})^{\mu(Q/d)} >\frac14\,2^{\varphi(Q)}.$$

To see the strict lower bound, omit the reciprocal factors and bound the remaining product below by the product over all positive integers. Its first two factors are $1/2$ and $3/4$; its tail is at least $1-\sum_{d\ge3}2^{-d}=3/4$. Their product is $9/32>1/4$. This proves (3). $\square$

This proof does not invoke a primitive-divisor existence theorem. It identifies and removes the possible exceptional cyclotomic prime directly.

Lemma 2. If $D_Q$ is the positive reduced denominator of $M_Q$, then

$$G_Q^2\mid D_Q, \qquad D_Q>\frac{2^{2\varphi(Q)-4}}{P^2}. \tag{7}$$

Proof. Let $\ell\mid G_Q$. Its order is $Q$, so it divides $2^Q-1$ and divides no $2^d-1$ for a proper divisor $d\mid Q$. Consequently only the $d=Q$ summand of $M_Q$ has a denominator divisible by $\ell$. Its numerator $\mu(Q)2^Q$ is an $\ell$-adic unit. Therefore

$$v_\ell(M_Q)=-2v_\ell(2^Q-1)=-2v_\ell(G_Q).$$

Adding the other $\ell$-integral summands cannot change that negative valuation. Multiplying the surviving prime powers proves the divisibility; (3) gives the inequality. $\square$

The readily available common denominator $(2^Q-1)^2$ is an upper bound. It plays no role in this reduced-denominator lower bound.

The obstruction after exact denominator clearing

Theorem 3. For every squarefree $Q$ with largest prime factor $P\ge5$,

$$D_Q R_Q> \frac{2^{2\varphi(Q)-q-4}}{P^2} \ge\frac{2^{2P-9}}{P^2}. \tag{8}$$

Consequently $D_QR_Q\to\infty$ along every sequence of squarefree $Q\to\infty$. For primorials the sharper lower bound $D_QR_Q>2^{2P-8}/P^2$ holds.

Proof. Combine (2) and (7). If $3\nmid Q$, then $q\le3$ and $\varphi(Q)\ge P-1$, whence $2\varphi(Q)-q-4\ge2P-9$. If $3\mid Q$, then $\varphi(Q)\ge2(P-1)$, since $3$ and $P$ are distinct prime divisors. Bertrand's postulate gives a prime between $P$ and $2P$, which cannot divide $Q$, so $q<2P$. Therefore $2\varphi(Q)-q-4>2P-8$. This case includes every primorial with $P\ge5$. Finally, only finitely many squarefree integers have all prime factors bounded by a fixed number, so $Q\to\infty$ forces $P\to\infty$.

For that external prime bound, the source states that for every integer $n\ge1$ there is a prime with $n<p\le2n$; at $n=P\ge5$ the upper endpoint is composite, so $q<2P$. See Erdős, Beweis eines Satzes von Tschebyschef (1932), pp. 197-198, section 5. Its source statement was inspected; the classical theorem is an external input, not a new proof claimed here. $\square$

If $S=a/b$ were rational, then

$$bD_Q(M_Q-S)=bD_QM_Q-aD_Q\in\mathbb Z_{>0}.$$

For each fixed $b$, (8) shows that these particular integers grow without bound. They cannot eventually lie strictly between zero and one. The same holds for any larger integer denominator clearing $M_Q$.

This closes only the use of $M_Q$ itself as an approximation with a small denominator-cleared error, for every choice of the finite prime set. It does not compare different $Q$ at a prescribed reduced denominator or claim that primorials are extremal in such a comparison. It does not exclude signed combinations of different $M_Q$, subtraction of further rational terms, other sieve weights, or multiplier approximants with independent cancellation. Those operations change both the reduced denominator and the full remainder, and require new estimates.