= Solution
Fix the noise convention explicitly: $p$ is the total probability of a nonidentity <Pauli error>. The memoryless qubit <quantum depolarizing channel> is
$$
\mathcal D_p(\rho)=(1-p)\rho+\frac p3(X\rho X+Y\rho Y+Z\rho Z),\qquad0\leq p\leq1,
$$
with independent identical use on successive qubits, namely $\mathcal D_p^{\otimes n}$ on an $n$-qubit block. Its normalized Kraus operators are $\sqrt{1-p}I,\sqrt{p/3}X,\sqrt{p/3}Y,\sqrt{p/3}Z$.
Write the <Bloch sphere> representation of a <density matrix> as $\rho=(I+\boldsymbol r\cdot\boldsymbol\sigma)/2$, with $|\boldsymbol r|\leq1$. Its eigenvalues are $(1\pm|\boldsymbol r|)/2$, proving exactly this positivity condition. Conjugation by one <Pauli operator> preserves that coordinate of the <Bloch vector> and reverses the other two. Summing the three conjugations gives
$$
X\rho X+Y\rho Y+Z\rho Z=2I-\rho,
$$
so
$$
\boxed{\boldsymbol r\longmapsto\eta\boldsymbol r,\qquad\eta=1-\frac{4p}{3}.}
$$
The Bloch ball contracts isotropically, collapses to the maximally mixed state at $p=3/4$, and contracts with reversal for $p>3/4$. That last interval still represents a physical Pauli mixture. If instead $q$ is the probability of replacing the state by $I/2$, then $\mathcal D(\rho)=(1-q)\rho+qI/2$, with $p=3q/4$ and $\eta=1-q$; the usual replacement-probability range $0\leq q\leq1$ stops at complete depolarization.
Let $h_2(x)=-x\log_2x-(1-x)\log_2(1-x)$, with endpoint limits. In bits per use, the unassisted product-state classical capacity and the <entanglement-assisted capacity of a qubit depolarizing channel> are
$$
\boxed{C_{\mathrm{prod}}(p)=1-h_2(2p/3),\qquad
C_E(p)=2-H_4(1-p,p/3,p/3,p/3),}
$$
where $H_4$ is Shannon entropy of the listed four probabilities. Equivalently $C_E=2+(1-p)\log_2(1-p)+p\log_2(p/3)$. The first quantity is a classical capacity with product quantum inputs and potentially collective output measurements, not the quantum capacity. These are the requested capacity formulae; a proof of their coding theorems is not needed here. The maximally entangled input gives a Bell-diagonal joint state with those four probabilities, explaining the entropy appearing in $C_E$. https://arxiv.org/abs/quant-ph/9904023[The original entanglement-assisted capacity calculation] gives the depolarizing-channel expression.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-32-capacities.png]
{title=Product-state and entanglement-assisted classical capacities under Pauli-error and replacement-probability conventions}
At $p=0$ the capacities are one and two. Both vanish at $p=3/4$. On the full Pauli-error interval they increase again beyond that point, reaching $1-h_2(2/3)$ and $2-\log_23$ at $p=1$; on the replacement-probability plot both decrease to zero at $q=1$. The two panels make this convention distinction explicit.
For the <high-noise capacity ratio for qubit depolarization>, put $\eta=1-4p/3\to0$. Taylor expansion of binary entropy around one half gives
$$
C_{\mathrm{prod}}=1-h_2\left(\frac{1-\eta}{2}\right)=\frac{\eta^2}{2\ln2}+O(\eta^4).
$$
The four probabilities for $C_E$ are $1/4+3\eta/4$ and three copies of $1/4-\eta/4$. For a perturbation $\delta_i$ of the uniform four-outcome distribution, with $\sum_i\delta_i=0$, entropy expansion gives
$$
2-H_4(1/4+\delta_i)=\frac1{2\ln2}\sum_i\frac{\delta_i^2}{1/4}+O(\|\delta\|^3).
$$
Here $\sum_i\delta_i^2=3\eta^2/4$, hence $C_E=3\eta^2/(2\ln2)+O(\eta^3)$. Therefore
$$
\boxed{\lim_{p\to3/4}\frac{C_E(p)}{C_{\mathrm{prod}}(p)}=3.}
$$
The reciprocal ratio is $1/3$, and the same enhancement factor occurs as $q\to1$ in the replacement convention.
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