Chapter 3 : Power Series. 1 Definitions and first properties P n Definition 1.1. A power series is a series of functions fn where fn : z 7! anz , (an) being a sequence of complex numbers. Depending on the cases, we will consider either the complex variable z, or the real variable x. Notations 1.2. For r ≥ 0, we will note ∆r = fz 2 C j jzj < rg, Kr = fz 2 C j jzj ≤ rg and Cr = fz 2 C j jzj = rg. P n Lemma 1.3. [Abel’s lemma] Let anz be a power series. We suppose that there ∗ n exists z0 2 C such that the sequence (anz0 ) is bounded. Then, for all r 2]0; jz0j[, P n anz normally converges on the compact Kr. Remark 1.4. P n • Note that it implies the absolute convergence on ∆jz0j, ie 8z 2 ∆jz0j, janz j converges. P n • Of course if we suppose janjr convergent, we directly have the normal n n convergence on Kr (cf. 8z 2 Kr, janz j ≤ janjr ). Proof. Let z be in Kr, we have n n n n n r r janz j ≤ janjr = janz0 j = O , jz0j jz0j which gives the result. P n Definition 1.5. We call the radius of convergence of the power series anz the number n + + R = supfr ≥ 0 j (anr ) boundedg 2 R = R [ f+1g. P n It will sometimes be noted RCV ( anz ). P n Theorem 1.6. Let R be the RCV of a power series anz . P n • For all r < R, anz normally converges on the compact Kr. n n1 • For all z such that jzj > R, anz 9 0. Remark 1.7. It implies the absolute convergence on ∆R. Proof. The Abel’s lemma gives the first point : 8r 2 [0;R[, 9r0 2]r; R] such that 0n (anr ) is bounded, which implies the normal convergence on Kr. For the second n n point, it’s the contraposition of anz ! 0 ) (anz ) bounded ) jzj ≤ R. 1 P n Corollary 1.8. With the same hypothesis, R = supfr ≥ 0 j anr convergesg = P n + inffr ≥ 0 j anr divergesg 2 R . 0 P n 00 Proof. Let’s note R = supfr ≥ 0 j anr convergesg and R = inffr ≥ P n 0 00 00 0 00 0 0 j anr divergesg. First, R ≤ R : if not, R < R and 9r 2]R ;R ] such P n n that anr converges, so we would have (anr ) bounded and convergence on ∆r (cf. 1.4), and thus R00 ≥ r, absurd. By the first point of the theorem, R0 ≥ R. By the second point, R00 ≤ R. So we have R ≤ R0 ≤ R00 ≤ R, which gives the result. Remark 1.9. • With the same kind of proof, one can show that we also have R = supfr ≥ n 0 j anr ! 0g. • To sum up, if we note C the domain of convergence of a power series which has a radius of convergence R, we have ∆R ⊂ C ⊂ KR and we have absolute convergence on ∆R. Definition 1.10. We call ∆R = fz 2 C j jzj < Rg the (open) disk of convergence. Remark 1.11. We can’t say anything a priori about the convergence of a power series on the circle CR, as we will see in the examples. Examples 1.12. • RCV (P zn) = 1 since the constant sequence (1) is bounded () RCV ≥ 1) P and 1 diverges () RCV ≤ 1). In fact there’s no point in C1 where there is oconvergence (jzj = 1 ) zn 9 0). • RCV (P zn=n) = 1 since (1=n) bounded () RCV ≥ 1) and P 1=n diverges () RCV ≤ 1). Here, the only point of C1 where the power series diverges is 1 : if z = eiθ 6= 1, P zn=n converges iff <(P zn=n) and =(P zn=n) converge, ie iff P cos(nθ)=n and P sin(nθ)=n converge. But we’ve already seen that the first one converges iff eiθ 6= 1, and the same proof shows that it’s the same for the second one. Exercise 1.13. [Hadamard theorem] Prove that this definition of the radius of convergence is equivalent to the first one : 1=n −1 R = (lim sup janj ) 2 Few methods to find the RCV P n Proposition 2.1. Let anz be a power series and z0 2 C. Then : P n P n • If anz0 converges but janz0 j diverges, then RCV = jz0j. P n n • Same conclusion if anz0 diverges but anz0 ! 0. Proof. For the first point, we have RCV ≥ jz0j (cf. 1.8), but we can’t have RCV > jz0j (cf. 1.6). The second point is a consequence of 1.8 and 1.9. P n P n Proposition 2.2. Let anz and bnz be two power series, and Ra;Rb their RCV. We have an = O(bn) ) Ra ≥ Rb. 2 n n P n Proof. Let z 2 ∆Rb , we have anz = O(bnz ) and jbnz j converges (cf. 1.6), P n so anz converges. By 1.8 we conclude Ra ≥ Rb. n n P n P n Remark 2.3. We can’t say that (anz = O(bnz ) and bnz converges) ) anz n + n converges because we’re not in the case bnz 2 R for n big enough (bnz 2 C). Corollary 2.4. With the same notations, we have an ∼ bn ) Ra = Rb. Proof. ∼ ) O. Proposition 2.5. Suppose an 6= 0 for n big enough. Then (with 1=0 = +1 and 1= + 1 = 0): an+1 + 1 9 lim = l 2 R ) RCV = : an l n+1 an+1z P n Proof. We have n ! ljzj. By De D’Alembert rule, jzj < 1=l ) anz anz P n converges, and RCV ≥ 1=l (cf. 1.8). Similarly, if jzj > 1=l, anz diverges, and RCV ≤ 1=l. P n Proposition 2.6. Let anz a power series and R its RCV. Then for all α 2 R P α n the RCV Rα of the power series n anz is also R. Proof. Let r < R and ρ 2]r; R[. We have n α n α r n α n n anr = n anρ ) (n anr ) bounded ) Rα ≥ R: ρ | {z } | {z } !0 !0 P n This is true for all anz , and for all α, so we also have, with β = −α, P β α n P α n R = RCV ( n (n anz )) ≥ RCV ( n anz ) = Rα. Examples 2.7. • By 2.5, RCV (P zn=n!) = +1. • By 2.5, RCV (P n!zn) = 0. P n 2 • By 2.6, RCV ( z =n ) = 1 and we have normal convergence on K1. P 2n n • We can abusively note z =5 the power series defined by a2n+1 = 0 and −n a2n = 5 for all n. But we can’t apply directly 2.5. However, it’s clear that we have convergence on ∆p and divergence on its complementary, so p 5 RCV = 5. P n P n Proposition 2.8. Let Ra and Rb be the RCV of anz and bnz . Then Ra+b = P n RCV ( (an + bn)z ) ≥ m = minfRa;Rbg, with equality if Ra 6= Rb. Moreover, on ∆m, we have P n P n P n (an + bn)z = anz + bnz . P n P n P Proof. For all z 2 ∆m, anz and bnz absolutely converges. Hence (an + n bn)z also does : Ra+b ≥ m and the additivity of limits of sequences gives the n additivity formula. If Ra < Rb, for all z 2 ∆Rb nKRa we have anz 9 0 and n n bnz ! 0, thus (an + bn)z 9 0, and Ra+b ≤ Ra = m. 3 P n P n P n P n n Example 2.9. Let anz = z and bnz = ((1=2) − 1)z , we have n Ra = 1 = Rb (use 2.5 for the second one). As an + bn = (1=2) , the domain of P n convergence of the (an + bn)z is clearly ∆2, so Ra+b = 2 > m. The nest result is obvious : ∗ P n P n Proposition 2.10. For all λ 2 C , anz and λanz have the same RCV R. Moreover, on ∆R, we have P n P n λanz = λ anz . P n P n Proposition 2.11. Let Ra and Rb be the RCV of anz and bnz . Then P n P n Ra?b = RCV (( anz ) ? ( bnz )) ≥ m = minfRa;Rbg. Moreover, on ∆m, we have P n P n P n P n ( anz ) ? ( bnz ) = ( anz )( bnz ). P n P n Proof. For all z 2 ∆m, janz j and jbnz j absolutely converges. Hence the P n P n Cauchy product ( janz j) ? ( jbnz j) converges (cf. ch1). But n n X k n−k X k n−k 8n; akz bn−kz ≤ jakz jjbn−kz j; k=0 k=0 so we get Ra?b ≥ m and the result given about the Cauchy product in chapter 1 gives the formula. Examples 2.12. P n • We don’t have the same result as for the addition if Ra 6= Rb : Let anz P n n+1 and bnz be defined by a0 = 1=2; b0 = −2 and an = −1=2 ; bn = −3 P n P n n+1 P n P n for n ≥ 1. We have anz = 1− n≥0 z =2 ; bnz = 1−3 n≥0 z , so Ra = 2 6= Rb = 1. We also have 1=2 z − 1 P a zn = 1 − = 8z 2 ∆ , n 1 − (z=2) z − 2 2 1 z − 2 and P b zn = 1 − 3 = 8z 2 ∆ .