Step 1: Focus directly on statement (B) using Taylor coefficients instead of the identity-theorem/pole argument. Expand $\dfrac{2n}{1+3n}=\dfrac{2}{3+1/n}$ as an asymptotic series in $1/n$ for large $n$: $\dfrac{2}{3+1/n}=\dfrac{2}{3}\cdot\dfrac{1}{1+\frac{1}{3n}}=\dfrac{2}{3}\left(1-\dfrac{1}{3n}+\dfrac{1}{9n^2}-\cdots\right)$.
Step 2: If an entire function $f$ satisfies $f(1/n)$ equal to this expression for every $n$, then since $f$ is analytic everywhere it has a Taylor series $f(z)=a_0+a_1z+a_2z^2+\cdots$ convergent for ALL $z\in\mathbb{C}$ (infinite radius of convergence, because $f$ is entire).
Step 3: Substituting $z=1/n$ into the Taylor series and matching term by term with the asymptotic expansion in Step 1 (valid since the values agree exactly at infinitely many points accumulating at $0$, which pins down every Taylor coefficient uniquely) forces $a_k=\dfrac{2(-1)^k}{3^{k+1}}$ for every $k\ge0$.
Step 4: But the power series $\sum a_kz^k$ with $a_k=\dfrac{2(-1)^k}{3^{k+1}}$ has $|a_k|^{1/k}\to\dfrac13$ as $k\to\infty$, so its radius of convergence is $R=\dfrac{1}{1/3}=3$, a FINITE number. This contradicts the requirement that $f$, being entire, has a Taylor series with infinite radius of convergence.
Step 5: The contradiction shows no such entire $f$ can exist, confirming statement (B) is false.\[\boxed{\text{Option (B) is false}}\]