Course 214 Section 2: Infinite Series Second Semester 2008

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1 Course 214 Section 2: Infinite Series Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins Contents 2 Infinite Series The Comparison Test and Ratio Test Absolute Convergence The Cauchy Product of Infinite Series Uniform Convergence for Infinite Series Power Series The Exponential Function i

2 2 Infinite Series An infinite series is the formal sum of the form a 1 + a 2 + a 3 +, where each number a n is real or complex. Such a formal sum is also denoted by + a n. Sometimes it is appropriate to consider infinite series + n=m a n of the form a m + a m+1 + a m+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 + a n is said to converge to some complex number s if and only if, given any ε > 0, there exists some natural number N m such that a n s < ε for all natural numbers m satisfying m N. If the infinite series + a n converges to s then we write + a n = s. An infinite series is said to be divergent if it is not convergent. For each natural number m, the mth partial sum s m of the infinite series + a n is given by s m = a 1 + a a m. Note that + a n converges to some complex number s if and only if s m s as m +. The following proposition therefore follows immediately on applying the results of Proposition 1.2. Proposition 2.1 Let + a n and + b n be convergent infinite series. Then + a n +b n is convergent, and + λ + a n for any complex number λ. a n +b n = + If + a n is convergent then a n 0 as n +. Indeed lim a n = n + where s m = m lim s n s n 1 = n + + a n + b n. Also + λa n = lim s n lim s n 1 = s s = 0, n + n + a n and s = + a n = lim m + s m. However the condition that a n 0 as n + is not in itself sufficient to ensure convergence. For example, the series + 1/n is divergent. 25

3 Proposition 2.2 Let a 1, a 2, a 3, a 4,... be an infinite sequence of real numbers. Suppose that a n 0 for all n. Then + a n is convergent if and only if there exists some real number C such that a 1 + a a n C for all n. Proof The sequence s 1, s 2, s 3,... of partial sums of the series + a n is nondecreasing, since a n 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 + z z m = 1 zm+1. 1 z To see this, multiply both sides of the equation by 1 z. Thus if z < 1 then lim 1 + z + m + z z m = 1 1 z. 2.1 The Comparison Test and Ratio Test Proposition 2.3 An infinite series + a n of real or complex numbers is convergent if and only if, given any ε > 0, there exists some natural number N with the property that a m + a m a m+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 a n b n for all n, where a n is complex, b n is real, and + b n is convergent. Then + is convergent. 26 a n

4 Proof Let ε > 0 be given. Then there exists some natural number N such that b m + b m b m+k < ε for all m and k satisfying m N and k 0. But then a m + a m a m+k a m + a m a m+k b m + b m b m+k < ε when m N and k 0. Thus + a n is convergent, by Proposition 2.3. Let us apply the Comparison Test in the case when a n and b n are nonnegative real numbers satisfying 0 a n b n for all n. If + b n is convergent, then so is + a n. Thus if + a n is divergent then so is + b n. These results also follow directly from Proposition 2.2. Example Comparison with the geometric series shows that the infinite series + z n /n is convergent whenever z < 1. Proposition 2.5 Ratio Test Let a 1, a 2, a 3... be complex numbers. Suppose that r = lim exists and satisfies r < 1. Then + a n is conver- a n+1 n + a n gent. Proof Choose ρ satisfying r < ρ < 1. Then there exists some natural number N such that a n+1 /a n < ρ for all n N. Let a1 K = maximum ρ, a 2 ρ, a 3 2 ρ,... a N. 3 ρ N Now a n+1 ρ a n whenever n N. Therefore a n ρ n N a N Kρ n whenever n N. But the choice of K also ensures that a n Kρ n when n < N. Moreover + Kρ n converges, since ρ < 1. The desired result therefore follows on applying the Comparison Test Proposition z n Example Let z be a complex number. Then converges for all values n! of z. For if a n = z n /n! then a n+1 /a n = z/n + 1, and hence a n+1 /a n 0 as n +. The result therefore follows on applying the Ratio Test. 27

5 Let + a n be an infinite series for which r = lim a n+1/a n is well-defined. n + The series clearly diverges if r > 1, since a n 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 Definition An infinite series + a n is said to be absolutely convergent if the infinite series + a n 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. Proposition 2.6 An infinite series + is absolutely convergent if and a n only if, given any ε > 0, there exists some natural number N such that a m + a m a m+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 + a n and + b n is defined to be the series + c n, where c n = n a j b n j = a 0 b n + a 1 b n 1 + a 2 b n a n 1 b 1 + a n b 0. j=0 The convergence of + a n and + b n is not in itself sufficient to ensure the convergence of the Cauchy product of these series. Convergence is however assured provided that the series + a n and + b n are absolutely convergent. 28

6 Theorem 2.7 The Cauchy product + c n of two absolutely convergent infinite series + a n and + b n is absolutely convergent, and c n = a n b n. Proof For each non-negative integer m, let S m = {j, k Z Z : 0 j m, 0 k m}, T m = {j, k Z Z : j 0, k 0, 0 j + k m}. Now m c n = m m a j b k and a n b n = a j b k. Also j,k T m j,k S m m c n + + a j b k a j b k a n b n, j,k T m j,k S m since c n n a j b n j and the infinite series + a n and + b n are absolutely j=0 convergent. It follows from Proposition 2.2 that the Cauchy product + is absolutely convergent, and is thus convergent. Moreover 2m m m c n a n b n = a j b k j,k T 2m \S m a j b k a j b k j,k T 2m \S m j,k S 2m \S m 2m 2m m m = a n b n a n b n, since S m T 2m S 2m. But 2m 2m a n b n lim m + = + + a n b n m m = lim a n b n, m + 29 c n

7 since the infinite series + a n and + b n are absolutely convergent. It follows that and hence as required. lim m + + c n = 2m m m c n a n b n = 0, lim m + 2m + + c n = a n b n, 2.4 Uniform Convergence for Infinite Series Let f 1, f 2, f 2,... be complex-valued functions defined over a subset D of C. The infinite series series + f n z is said to converge uniformly on D to some function s if, given any ε > 0, there exists some natural number N which n does not depend on the value of z such that sz f m z < ε whenever m=0 z D and n N. Note that an infinite series + f n z of functions converges uniformly if and only if the partial sums of this series converge uniformly. It follows immediately from Theorem 1.20 that if the functions f n are continuous on D, and if the series + f n z converges uniformly on D to some function, then that function is also continuous on D. Proposition 2.8 The Weierstrass M-Test Let D be a subset of C and let f 1, f 2, f 3,... be a sequence of functions from D to C, let M 1, M 2, M 3,... be non-negative real numbers satisfying f n z M n for all natural numbers n and z D. Suppose that + M n converges. Then + f n z converges absolutely and uniformly on D. Proof It follows immediately from the Comparison Test Proposition 2.4 that the series + f n z is absolutely convergent for all z D. We must show that the convergence is uniform. Let ε > 0 be given. Then there exists some natural number N such that + n=n M n < 1ε, since + M 2 n converges. Now if m and k are integers 30

8 satisfying m N and k 0 then m+k m f n z f n z = m+k n=m+1 f n z m+k n=m+1 M n for any z D. On taking the limit as k +, we see that + m f n z f n z 1ε < ε 2 + n=n M n < 1 2 ε 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 + a n z z 0 n, where the coefficients a 0, a 1, a 2,... are complex numbers. Definition Let + a n z z 0 n be a power series centred on some complex number z 0. Suppose that the set of complex numbers z for which the power series converges is bounded. Then the radius of convergence R 0 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 z 0 R 0. The circle {z C : z z 0 = R 0 } is then referred to as the circle of convergence of the power series. We set R 0 = + if the set of complex numbers z for which the power series converges is unbounded. Theorem 2.9 Let + a n z z 0 n be a power series with radius of convergence R 0, and let sz denote the sum of the power series at those complex numbers z at which the series converges. i If R 0 = + then sz is a continuous function of z defined over the entire complex plane C. ii If R 0 < + then sz is a continuous function of z defined over the whole of the disk {z C : z z 0 < R 0 } bounded by the circle of convergence of the power series. 31

9 Proof Let z 1 be any complex number satisfying z 1 z 0 < R 0. Then we can choose R such that z 1 z 0 < R < R 0 and R < +. Now it follows from the definition of the radius of convergence that there exists some complex number w such that R < w < R 0 and + a n w n converges. Choose some positive real number A with the property that a n w n A for all n, and set ρ = R/ w and M n = Aρ n. If z z 0 < R then a n z z 0 n a n R n Aρ n = M n for all n. Also + M n converges to A/1 ρ. Thus we can apply the Weierstrass M-Test Proposition 2.8 to deduce that the power series + a n z z 0 n converges uniformly on the disk {z C : z z 0 < R} of radius R about z 0. 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 z 1. We deduce that the function s is continuous throughout the complex plane when R 0 = +, and is continuous inside the circle of convergence when R 0 < +, 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 expz = A straightforward application of the Ratio Test shows that radius of convergence of the power series defining expz is infinite, and it therefore follows from Theorem 2.9 that the exponential function is continuous on C. + z n n!. Lemma 2.10 The exponential function has the property that for all complex numbers z and w. expz + w = expz expw. Proof Let z and w be complex numbers. The infinite series defining expz and expw are absolutely convergent. It therefore follows from Theorem

10 that the value of expz expw is the sum of the Cauchy product of the infinite series defining expz and expw, and therefore where j=0 expz expw = c n = n j=0 z j j! + c n w n j n j! On applying the Binomial Theorem, we see that c n = 1 n n! n! j!n j! zj w n j = 1 n n z j w n j = 1 n! j n! z + wn, j=0 and thus expz expw = expz + w, as required. Let z be a complex number. Then expz exp z = expz z = exp0 = 1. It follows that expz 0 and exp z = 1/ expz 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 = 1 n z 2n+1, cos z = 2n + 1! 1 n z 2n. 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 cos z = 1 expiz + exp iz, 2 sin z = 1 expiz exp iz, 2i cos z + i sin z = expiz, 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 expz = expx + iy = expx expiy = e x cos y + i sin y. 33

11 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 = rcos θ + i sin θ. Moreover there exists a real number x such that r = e x. This real number x is the natural logarithm log e r of r. Then w = expx + iθ. Lemma 2.11 Let z and w be complex numbers. Then expz = expw if and only if w = z + 2πin for some integer n. Proof If w = z + 2πin for some integer n then expw = expz exp2πin = expzcos 2πn + i sin 2πn = expz. Conversely suppose that expw = expz. Let w z = u+iv, where u, v R. Then e u cos v + i sin v = expw z = expw expz 1 = 1. Taking the modulus of both sides, we see that e u = 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 D 0 = C \ {x R : x 0}. Then there exists a unique continuous function log: D 0 C characterized by the properties that log1 = 0, Im logz < π and explogz = z for all z D 0. This function has the property that logre iθ = log e r + iθ for all real numbers r and θ satisfying r > 0 and π < θ < π, where log e r denotes the natural logarithm of the real number r. Proof Given any complex number w belonging to the open set D 0 there exist unique real numbers r and θ for which r > 0, π < θ < π and re iθ = w. It follows that there is a well-defined function log: D 0 C such that logre iθ = log e r+iθ for all real numbers real numbers r and θ satisfying r > 0 and π < θ < π, where log e r, the natural logarithm of r, is the unique real number t satisfying e t = r. Moreover explogre iθ = explog e r expiθ = re iθ for all real numbers r and θ with r > 0. It follows that explogz = z for all z D 0. 34

12 Let x and y be real numbers. Then 1 log 2 ex 2 + y 2 + i arccos 1 logx + iy = log 2 ex 2 + y 2 i arccos 1 log 2 ex 2 + y 2 + i arcsin x x2 + y 2 x x2 + y 2 y x2 + y 2 if y > 0, if y < 0, if x > 0, 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 log e, 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 We conclude that the restrictions of log: D 0 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 D 0 is the union of these three open sets. It follows that log: D 0 C is continuous throughout D 0. A straightforward application of Lemma 2.11 then shows that this function log is the unique function from D 0 to C with the required properties. The function log: D 0 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 explz = 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 D w, w = {z C : z w < w }. Then there exists a continuous function F w : D w, w C with the property that expf w z = z for all z D w, w. Proof Let D 0 = C \ {x R : x 0}, let log: D 0 C be the principal branch of the logarithm function, and let ζ be a complex number satisfying exp ζ = w. Then z/w D 0 for all z D w, w. A function F w : D w, w C with the required properties may therefore be obtained on defining F w z = ζ + logz/w for all z D w, w. 35