By Larry Joel Goldstein

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Let a > 0. If for real s > 0 F(s) =A f(l +a) (1 + r(s)), s0t where lr(s)I :::; Cose and A> 0, Co > 0, and€> 0 are constants, then fox f(t) dr(t) where Ip( x) I :::; Ci/ ln x, = AxOt(l + p(x)), C 1 > 0 being a constant. We will show how this theorem contains the Hardy-Littlewood theorem (with remainder term) for a power series. Suppose that as z -+ 1 a series with nonnegative coefficients an satisfies a relation ~ anzn = where e > 0. Then as s -+ (l ! z)0t (1+0((1 - zt)), 0 L 00 1 ane-ns = 0 (1 + O(sc)), 8 n=O where€> 0.

This proves the lemma. LEMMA 11. Suppose 11",,(x) = E~ akxk is a polynomial of degree v with real coefficients such that 17r,,(x)I $ M on the interval [O, l]. Then its coefficients satisfy the estimate where C is an absolute constant. PROOF. In this lemma the letter C will stand for absolute constants, not necessarily equal to each other. Consider the v + 1 points Xi = i / v, 48 1. FACTS FROM ANALYSIS i = 0, 1, ... , v, in [O, l]. Since a polynomial of degree vis determined by its values at L' + 1 points, it follows from the Lagrange interpolation formula that 11",,,(x) L 11",,,(xi) (xi = i=O (x - xo) · · · (x - Xi-1)(x - Xi+i) · · · (x - x,,,) ( · xo) ···(xi - Xi-1)(xi - Xi+i) · · · Xi - x,,,) v Let

TAUBERIAN THEOREMS FOR POWER SERIES 29 Postnikov [115] proved a complex analogue of the Hardy-Littlewood Tauberian theorem. Suppose a power series f(z) = E;;" anzn has radius of convergence equal to l. Let z = reie (r = lzl). Assume that f(z) = 1/(1- z) + 0(1) for IOI :::; c < 1r. Then Lan= N + O(lnN). THEOREM. n5,N Comparing Postnikov's theorem with Fatou's, we see that imposing conditions in the complex domain leads (in the given situation) to a significant improvement in the remainder term. There is a complex Tauberian theorem for Dirichlet integrals, namely Ikehara's theorem, that is useful in analytic number theory.