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Course 214 Section 2: Infinite Series Second Semester 2008 David R. Wilkins c David R. Wilkins 1989–2008 Copyright Contents 2 Infinite Series 2.1 The Comparison Test and Ratio Test . 2.2 Absolute Convergence . . . . . . . . . 2.3 The Cauchy Product of Infinite Series . 2.4 Uniform Convergence for Infinite Series 2.5 Power Series . . . . . . . . . . . . . . . 2.6 The Exponential Function . . . . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 26 28 28 30 31 32 2 Infinite Series An infinite series is the formal sum of the form a1 + a2 + a3 + · · ·, where each +∞ P number an is real or complex. Such a formal sum is also denoted by an . n=1 Sometimes it is appropriate to consider infinite series +∞ P an of the form n=m am + am+1 + am+2 + · · ·, where m ∈ Z. Clearly results for such sequences may be deduced immediately from corresponding results in the case m = 1. Definition An infinite series +∞ P an is said to converge to some complex n=1 number s if and only if, given any ε > 0, there exists some natural number N m X an − s < ε for all natural numbers m satisfying m ≥ N . If such that n=1 +∞ +∞ P P the infinite series an converges to s then we write an = s. An infinite n=1 n=1 series is said to be divergent if it is not convergent. For each natural number m, the mth partial sum sm of the infinite se+∞ +∞ P P ries an is given by sm = a1 + a2 + · · · + am . Note that an converges n=1 n=1 to some complex number s if and only if sm → s as m → +∞. The following proposition therefore follows immediately on applying the results of Proposition 1.2. +∞ P Proposition 2.1 Let an and n=1 +∞ P λ bn be convergent infinite series. Then n=1 (an +bn ) is convergent, and n=1 +∞ P +∞ P +∞ X (an +bn ) = n=1 +∞ X n=1 an + +∞ X bn . Also n=1 +∞ P (λan ) = n=1 an for any complex number λ. n=1 If +∞ P an is convergent then an → 0 as n → +∞. Indeed n=1 lim an = lim (sn − sn−1 ) = lim sn − lim sn−1 = s − s = 0, n→+∞ where sm = n→+∞ m P an and s = n=1 n→+∞ +∞ P n→+∞ an = limm→+∞ sm . However the condition n=1 that an → 0 as n → +∞ is not in itself sufficient to ensure convergence. For +∞ P example, the series 1/n is divergent. n=1 25 Proposition 2.2 Let a1 , a2 , a3 , a4 , . . . be an infinite sequence of real num+∞ P an is convergent if and only if bers. Suppose that an ≥ 0 for all n. Then n=1 there exists some real number C such that a1 + a2 + · · · + an ≤ C for all n. Proof The sequence s1 , s2 , s3 , . . . of partial sums of the series +∞ P an is non- n=1 decreasing, since an ≥ 0 for all n. The result therefore is a consequence of the fact that a non-decreasing sequence of real numbers is convergent if and only if it is bounded above (see Theorem 1.3). Example Let z be a complex number. The infinite series 1+z +z 2 +z 3 +· · · is referred to as the geometric series. This series is clearly divergent whenever |z| ≥ 1, since z n does not converge to 0 as n → +∞. We claim that the series converges to 1/(1 − z) whenever |z| < 1. Now 1 + z + z2 + . . . zm = 1 − z m+1 . 1−z (To see this, multiply both sides of the equation by 1 − z.) Thus if |z| < 1 then 1 lim (1 + z + z 2 + . . . z m ) = . m→+∞ 1−z 2.1 The Comparison Test and Ratio Test Proposition 2.3 An infinite series +∞ P an of real or complex numbers is con- n=1 vergent if and only if, given any ε > 0, there exists some natural number N with the property that |am + am+1 + · · · + am+k | < ε for all m and k satisfying m ≥ N and k ≥ 0. Proof The stated criterion is equivalent to the condition that the sequence of partial sums of the series be a Cauchy sequence. The required result thus follows immediately from Cauchy’s Criterion for convergence (Theorem 1.9). Proposition 2.4 (Comparison Test) Suppose that 0 ≤ |an | ≤ bn for all n, +∞ +∞ P P where an is complex, bn is real, and bn is convergent. Then an is convergent. n=1 26 n=1 Proof Let ε > 0 be given. Then there exists some natural number N such that bm + bm+1 + · · · + bm+k < ε for all m and k satisfying m ≥ N and k ≥ 0. But then |am + am+1 + · · · + am+k | ≤ |am | + |am+1 | + · · · + |am+k | ≤ bm + bm+1 + · · · + bm+k < ε when m ≥ N and k ≥ 0. Thus +∞ P an is convergent, by Proposition 2.3. n=1 Let us apply the Comparison Test in the case when an and bn are non+∞ P bn is convergent, negative real numbers satisfying 0 ≤ an ≤ bn for all n. If then so is +∞ P an . Thus if n=1 +∞ P an is divergent then so is n=1 n=1 +∞ P bn . These results n=1 also follow directly from Proposition 2.2. Example Comparison with the geometric series shows that the infinite series +∞ P n z /n is convergent whenever |z| < 1. n=1 Proposition 2.5 (Ratio Test) Let a1 , a2 , a3 . . . be complex numbers. Sup+∞ P an+1 exists and satisfies |r| < 1. Then an is converpose that r = lim n→+∞ an n=1 gent. Proof Choose ρ satisfying |r| < ρ < 1. Then there exists some natural number N such that |an+1 /an | < ρ for all n ≥ N . Let |a1 | |a2 | |a3 | |aN | , ,... N . K = maximum , ρ ρ2 ρ3 ρ Now |an+1 | ≤ ρ|an | whenever n ≥ N . Therefore |an | ≤ ρn−N |aN | ≤ Kρn whenever n ≥ N . But the choice of K also ensures that |an | ≤ Kρn when n < +∞ P N . Moreover Kρn converges, since ρ < 1. The desired result therefore n=1 follows on applying the Comparison Test (Proposition 2.4). Example Let z be a complex number. Then +∞ n X z converges for all values n! n=0 of z. For if an = z n /n! then an+1 /an = z/(n + 1), and hence an+1 /an → 0 as n → +∞. The result therefore follows on applying the Ratio Test. 27 Let +∞ P an be an infinite series for which r = lim an+1 /an is well-defined. n→+∞ n=1 The series clearly diverges if |r| > 1, since |an | increases without limit as n → +∞. If however |r| = 1 then the Ratio Test is of no help in deciding whether or not the series converges, and one must try other more sensitive tests. 2.2 Absolute Convergence P+∞ Definition An infinite series n=1 an is said to be absolutely convergent if P+∞ the infinite series n=1 |an | is convergent. A convergent series which is not absolutely convergent is said to be conditionally convergent. An absolutely convergent infinite series is convergent, and the sum of any two absolutely convergent series is itself absolutely convergent. These results follow on applying the Comparison Test (Proposition 2.4). Moreover the following criterion for absolute convergence follows directly from Proposition 2.3. +∞ P Proposition 2.6 An infinite series an is absolutely convergent if and n=1 only if, given any ε > 0, there exists some natural number N such that |am | + |am+1 | + · · · + |am+k | < ε for all m and k satisfying m ≥ N and k ≥ 0. Many of the tests for convergence described above do in fact test for absolute convergence; these include the Comparison Test and the Ratio Test. 2.3 The Cauchy Product of Infinite Series The Cauchy product of two infinite series +∞ P an and n=0 the series +∞ P +∞ P bn is defined to be n=0 cn , where n=0 cn = n X aj bn−j = a0 bn + a1 bn−1 + a2 bn−2 + · · · + an−1 b1 + an b0 . j=0 The convergence of +∞ P n=0 an and +∞ P bn is not in itself sufficient to ensure the n=0 convergence of the Cauchy product of these series. Convergence is however +∞ +∞ P P bn are absolutely convergent. assured provided that the series an and n=0 n=0 28 Theorem 2.7 The Cauchy product +∞ P cn of two absolutely convergent infi- n=0 nite series +∞ P an and n=0 +∞ P bn is absolutely convergent, and n=0 +∞ X +∞ X cn = n=0 ! +∞ X an n=0 ! bn . n=0 Proof For each non-negative integer m, let Sm = {(j, k) ∈ Z × Z : 0 ≤ j ≤ m, 0 ≤ k ≤ m}, Tm = {(j, k) ∈ Z × Z : j ≥ 0, k ≥ 0, 0 ≤ j + k ≤ m}. m m m P P P P P Now aj bk . Also cn = aj bk and an bn = n=0 m X n=0 (j,k)∈Tm X |cn | ≤ n=0 n P since |cn | ≤ X |aj ||bk | ≤ (j,k)∈Tm n=0 (j,k)∈Sm +∞ X |aj ||bk | ≤ ! +∞ X |an | n=0 (j,k)∈Sm |aj ||bn−j | and the infinite series +∞ P |bn | , n=0 +∞ P an and n=0 j=0 ! bn are absolutely n=0 convergent. It follows from Proposition 2.2 that the Cauchy product +∞ P n=0 is absolutely convergent, and is thus convergent. Moreover ! m ! 2m m X X X bn cn − an n=0 n=0 n=0 X = aj bk (j,k)∈T2m \Sm X X ≤ |aj bk | ≤ |aj bk | (j,k)∈T2m \Sm = 2m X ! |an | n=0 (j,k)∈S2m \Sm 2m X ! |bn | n=0 − n=0 since Sm ⊂ T2m ⊂ S2m . But ! 2m ! 2m X X lim |an | |bn | = m→+∞ m X n=0 29 |an | n=0 +∞ X ! |an | n=0 = ! lim m→+∞ m X ! |bn | , n=0 +∞ X ! |bn | n=0 m X n=0 ! |an | m X n=0 ! |bn | , cn since the infinite series +∞ P an and n=0 that 2m X lim m→+∞ and hence +∞ X bn are absolutely convergent. It follows n=0 m X cn − n=0 ! an n=0 cn = lim m→+∞ n=0 +∞ P 2m X cn = n=0 m X !! bn = 0, n=0 +∞ X n=0 ! an +∞ X ! bn , n=0 as required. 2.4 Uniform Convergence for Infinite Series Let f1 , f2 , f2 , . . . be complex-valued functions defined over a subset D of C. +∞ P The infinite series series fn (z) is said to converge uniformly on D to some n=1 function s if, given any ε > 0, there exists some natural number N (which n X fm (z) < ε whenever does not depend on the value of z) such that s(z) − m=0 z ∈ D and n ≥ N . +∞ P Note that an infinite series fn (z) of functions converges uniformly if n=0 and only if the partial sums of this series converge uniformly. It follows immediately from Theorem 1.20 that if the functions fn are continuous on +∞ P D, and if the series fn (z) converges uniformly on D to some function, n=0 then that function is also continuous on D. Proposition 2.8 (The Weierstrass M -Test) Let D be a subset of C and let f1 , f2 , f3 , . . . be a sequence of functions from D to C, let M1 , M2 , M3 , . . . be non-negative real numbers satisfying |fn (z)| ≤ Mn for all natural numbers n +∞ +∞ P P and z ∈ D. Suppose that Mn converges. Then fn (z) converges abson=1 n=1 lutely and uniformly on D. Proof It follows immediately from the Comparison Test (Proposition 2.4) +∞ P that the series fn (z) is absolutely convergent for all z ∈ D. We must n=1 show that the convergence is uniform. Let ε > 0 be given. Then there exists some natural number N such +∞ P P 1 that +∞ M < ε, since Mn converges. Now if m and k are integers n n=N 2 n=1 30 satisfying m ≥ N and k ≥ 0 then m+k +∞ m m+k m+k X X X X X Mn ≤ Mn < 21 ε fn (z) ≤ fn (z) − fn (z) = n=m+1 n=1 n=m+1 n=1 n=N for any z ∈ D. On taking the limit as k → +∞, we see that +∞ m X X fn (z) − fn (z) ≤ 21 ε < ε n=1 n=1 for all z ∈ D and m ≥ N . However N has been chosen independently of z. Thus the infinite series converges uniformly on D, as required. 2.5 Power Series A power series is an infinite series of the form +∞ P an (z − z0 )n , where the n=0 coefficients a0 , a1 , a2 , . . . are complex numbers. Definition Let +∞ P an (z − z0 )n be a power series centred on some complex n=0 number z0 . Suppose that the set of complex numbers z for which the power series converges is bounded. Then the radius of convergence R0 of the power series is defined to be the smallest non-negative real number with the property that every complex number z for which the power series converges satisfies |z − z0 | ≤ R0 . The circle {z ∈ C : |z − z0 | = R0 } is then referred to as the circle of convergence of the power series. We set R0 = +∞ if the set of complex numbers z for which the power series converges is unbounded. Theorem 2.9 Let +∞ P an (z − z0 )n be a power series with radius of conver- n=0 gence R0 , and let s(z) denote the sum of the power series at those complex numbers z at which the series converges. (i) If R0 = +∞ then s(z) is a continuous function of z defined over the entire complex plane C. (ii) If R0 < +∞ then s(z) is a continuous function of z defined over the whole of the disk {z ∈ C : |z − z0 | < R0 } bounded by the circle of convergence of the power series. 31 Proof Let z1 be any complex number satisfying |z1 − z0 | < R0 . Then we can choose R such that |z1 −z0 | < R < R0 and R < +∞. Now it follows from the definition of the radius of convergence that there exists some complex num+∞ P ber w such that R < |w| < R0 and an wn converges. Choose some positive n=0 real number A with the property that |an wn | ≤ A for all n, and set ρ = R/|w| n and Mn = Aρ . If |z − z0 | < R then |an (z − z0 )n | ≤ |an |Rn ≤ Aρn = Mn for P+∞ all n. Also n=0 Mn converges to A/(1 − ρ). Thus we can apply the Weier+∞ P strass M -Test (Proposition 2.8) to deduce that the power series an (z−z0 )n n=0 converges uniformly on the disk {z ∈ C : |z − z0 | < R} of radius R about z0 . It then follows from Theorem 1.20 that the restriction of the function s to this disk is continuous on the disk, and, in particular, is continuous around z1 . We deduce that the function s is continuous throughout the complex plane when R0 = +∞, and is continuous inside the circle of convergence when R0 < +∞, as required. A power series with finite radius of convergence will converge everywhere within its circle of convergence, and will diverge everywhere outside this circle. However Theorem 2.9 provides no information concerning the behaviour of the power series on the circle of convergence itself. 2.6 The Exponential Function Definition The exponential function exp: C → C is defined for all complex numbers z by the formula exp(z) = +∞ n X z n=0 n! . A straightforward application of the Ratio Test shows that radius of convergence of the power series defining exp(z) is infinite, and it therefore follows from Theorem 2.9 that the exponential function is continuous on C. Lemma 2.10 The exponential function has the property that exp(z + w) = exp(z) exp(w). for all complex numbers z and w. Proof Let z and w be complex numbers. The infinite series defining exp(z) and exp(w) are absolutely convergent. It therefore follows from Theorem 2.7 32 that the value of exp(z) exp(w) is the sum of the Cauchy product of the infinite series defining exp(z) and exp(w), and therefore exp(z) exp(w) = +∞ X cn n=0 where cn = n X zj j=0 wn−j j! (n − j)! On applying the Binomial Theorem, we see that n n n! 1 1 X 1 X n j n−j j n−j z w = (z + w)n , cn = z w = n! j=0 j!(n − j)! n! j=0 j n! and thus exp(z) exp(w) = exp(z + w), as required. Let z be a complex number. Then exp(z) exp(−z) = exp(z − z) = exp(0) = 1. It follows that exp(z) 6= 0 and exp(−z) = 1/ exp(z) for all complex numbers z. Let sin: C → C and cos: C → C be the functions defined as sums of power series according to the formulae sin z = ∞ X (−1)n z 2n+1 n=0 (2n + 1)! , cos z = ∞ X (−1)n z 2n n=0 (2n)! . If z is real then the power series appearing on the right hand side of these identities correspond to the Taylor expansions of the standard sine and cosine functions. A routine application of the Ratio Test shows that these power series both have infinite radius of convergence, and therefore define continuous functions defined over the whole complex plane. These continuous functions therefore constitute a natural extension to the complex plane of the standard sine and cosine functions of basic trigonometry. On comparing power series, we see that 1 cos z = (exp(iz) + exp(−iz)), 2 cos z + i sin z = exp(iz), sin z = 1 (exp(iz) − exp(−iz)), 2i cos z − i sin z = exp(−iz) for all complex numbers z. Thus if z = x + iy, where x and y are real numbers, then exp(z) = exp(x + iy) = exp(x) exp(iy) = ex (cos y + i sin y). 33 The exponential map exp: C → C \ {0} is surjective. For, given any complex number w, there exist real numbers r and θ such that r > 0 and w = r(cos θ + i sin θ). Moreover there exists a real number x such that r = ex . (This real number x is the natural logarithm loge r of r.) Then w = exp(x + iθ). Lemma 2.11 Let z and w be complex numbers. Then exp(z) = exp(w) if and only if w = z + 2πin for some integer n. Proof If w = z + 2πin for some integer n then exp(w) = exp(z) exp(2πin) = exp(z)(cos 2πn + i sin 2πn) = exp(z). Conversely suppose that exp(w) = exp(z). Let w−z = u+iv, where u, v ∈ R. Then eu (cos v + i sin v) = exp(w − z) = exp(w) exp(z)−1 = 1. Taking the modulus of both sides, we see that eu = 1, and thus u = 0. Also cos v = 1 and sin v = 0, and therefore v = 2πn for some integer n. The result follows. Proposition 2.12 Let D0 = C \ {x ∈ R : x ≤ 0}. Then there exists a unique continuous function log: D0 → C characterized by the properties that log(1) = 0, | Im log(z)| < π and exp(log(z)) = z for all z ∈ D0 . This function has the property that log(reiθ ) = loge r + iθ for all real numbers r and θ satisfying r > 0 and −π < θ < π, where loge r denotes the natural logarithm of the real number r. Proof Given any complex number w belonging to the open set D0 there exist unique real numbers r and θ for which r > 0, −π < θ < π and reiθ = w. It follows that there is a well-defined function log: D0 → C such that log(reiθ ) = loge r+iθ for all real numbers real numbers r and θ satisfying r > 0 and −π < θ < π, where loge r, the natural logarithm of r, is the unique real number t satisfying et = r. Moreover exp(log(reiθ )) = exp(loge r) exp(iθ) = reiθ for all real numbers r and θ with r > 0. It follows that exp(log(z)) = z for all z ∈ D0 . 34 Let x and y be real numbers. Then ! x 1 loge (x2 + y 2 ) + i arccos p if y > 0, 2 2 + y2 x ! x if y < 0, log(x + iy) = 21 loge (x2 + y 2 ) − i arccos p 2 + y2 x ! y 2 2 1 if x > 0, 2 loge (x + y ) + i arcsin p 2 x + y2 where arcsin: [−1, 1] → [−π/2, π/2] and arccos: [−1, 1] → [0, π] are the inverses of the restrictions of the sine and cosine functions to the intervals [−π/2, π/2] and [0, π] respectively. Now the functions loge , arcsin and arccos are continuous, since the inverse of any strictly increasing or strictly decreasing function defined over some interval in R is itself continuous. (see Corollary 1.19). We conclude that the restrictions of log: D0 → C to each of the open sets {z ∈ C : Im z > 0}, {z ∈ C : Im z < 0}, {z ∈ C : Re z < 0} is continuous, and the open set D0 is the union of these three open sets. It follows that log: D0 → C is continuous throughout D0 . A straightforward application of Lemma 2.11 then shows that this function log is the unique function from D0 to C with the required properties. The function log: D0 → C characterized by the properties set out in the statement of Proposition 2.12 is referred to as the principal branch of the logarithm function. It is impossible to define a continuous ‘logarithm’ function L: C \ {0} → C with the property that exp(L(z)) = z for all z ∈ C \ {0}. The best that one can achieve is to define continuous inverses of the exponential map over appropriately chosen open subsets of C \ {0}. Such functions are referred to as ‘branches of the logarithm function’. Corollary 2.13 Let w be a non-zero complex number, and let Dw,|w| = {z ∈ C : |z − w| < |w|}. Then there exists a continuous function Fw : Dw,|w| → C with the property that exp(Fw (z)) = z for all z ∈ Dw,|w| . Proof Let D0 = C \ {x ∈ R : x ≤ 0}, let log: D0 → C be the principal branch of the logarithm function, and let ζ be a complex number satisfying exp ζ = w. Then z/w ∈ D0 for all z ∈ Dw,|w| . A function Fw : Dw,|w| → C with the required properties may therefore be obtained on defining Fw (z) = ζ + log(z/w) for all z ∈ Dw,|w| . 35