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Course 214 Section 2: Infinite Series Second Semester 2008

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Course 214 Section 2: Infinite Series Second Semester 2008 Course 214 Section 2: Infinite Series Second Semester 2008 David R. Wilkins Copyright c David R. Wilkins 1989–2008 Contents 2 Infinite Series 25 2.1 The Comparison Test and Ratio Test . 26 2.2 Absolute Convergence . 28 2.3 The Cauchy Product of Infinite Series . 28 2.4 Uniform Convergence for Infinite Series . 30 2.5 Power Series . 31 2.6 The Exponential Function . 32 i 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 +∞ P Sometimes it is appropriate to consider infinite series 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. +∞ P Definition An infinite series 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 such that an − s < ε for all natural numbers m satisfying m ≥ N. If 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 fol- lowing proposition therefore follows immediately on applying the results of Proposition 1.2. +∞ +∞ P P Proposition 2.1 Let an and bn be convergent infinite series. Then n=1 n=1 +∞ +∞ +∞ +∞ +∞ P X X X P (an+bn) is convergent, and (an+bn) = an+ bn. Also (λan) = n=1 n=1 n=1 n=1 n=1 +∞ P λ an for any complex number λ. n=1 +∞ P If 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→+∞ n→+∞ n→+∞ n→+∞ m +∞ P P where sm = an and s = an = limm→+∞ sm. However the condition n=1 n=1 that an → 0 as n → +∞ is not in itself sufficient to ensure convergence. For +∞ example, the series P 1/n is divergent. n=1 25 Proposition 2.2 Let a1, a2, a3, a4,... be an infinite sequence of real num- +∞ P bers. Suppose that an ≥ 0 for all n. Then an is convergent if and only if n=1 there exists some real number C such that a1 + a2 + ··· + an ≤ C for all n. +∞ P Proof The sequence s1, s2, s3,... of partial sums of the series 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 +z2 +z3 +··· is referred to as the geometric series. This series is clearly divergent whenever |z| ≥ 1, since zn does not converge to 0 as n → +∞. We claim that the series converges to 1/(1 − z) whenever |z| < 1. Now 1 − zm+1 1 + z + z2 + . zm = . 1 − z (To see this, multiply both sides of the equation by 1 − z.) Thus if |z| < 1 then 1 lim (1 + z + z2 + . zm) = . m→+∞ 1 − z 2.1 The Comparison Test and Ratio Test +∞ P Proposition 2.3 An infinite series 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 (Theo- rem 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 n=1 n=1 convergent. 26 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 +∞ P when m ≥ N and k ≥ 0. Thus 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 negative real numbers satisfying 0 ≤ an ≤ bn for all n. If bn is convergent, n=1 +∞ +∞ +∞ P P P then so is an. Thus if an is divergent then so is bn. These results n=1 n=1 n=1 also follow directly from Proposition 2.2. Example Comparison with the geometric series shows that the infinite series +∞ P zn/n is convergent whenever |z| < 1. n=1 Proposition 2.5 (Ratio Test) Let a1, a2, a3 ... be complex numbers. Sup- +∞ an+1 P pose that r = lim exists and satisfies |r| < 1. Then an is conver- 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 |a | |a | |a | |a | K = maximum 1 , 2 , 3 ,... N . ρ ρ2 ρ3 ρN n−N n Now |an+1| ≤ ρ|an| whenever n ≥ N. Therefore |an| ≤ ρ |aN | ≤ Kρ n whenever n ≥ N. But the choice of K also ensures that |an| ≤ Kρ when n < +∞ N. Moreover P Kρn converges, since ρ < 1. The desired result therefore n=1 follows on applying the Comparison Test (Proposition 2.4). +∞ X zn Example Let z be a complex number. Then converges for all values n! n=0 n of z. For if an = z /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 +∞ P Let an be an infinite series for which r = lim an+1/an is well-defined. n=1 n→+∞ 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 Propo- sition 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 +∞ +∞ P P The Cauchy product of two infinite series an and bn is defined to be n=0 n=0 +∞ P the series cn, where n=0 n X cn = ajbn−j = a0bn + a1bn−1 + a2bn−2 + ··· + an−1b1 + anb0. j=0 +∞ +∞ P P The convergence of an and bn is not in itself sufficient to ensure the n=0 n=0 convergence of the Cauchy product of these series. Convergence is however +∞ +∞ P P assured provided that the series an and bn are absolutely convergent. n=0 n=0 28 +∞ P Theorem 2.7 The Cauchy product cn of two absolutely convergent infi- n=0 +∞ +∞ P P nite series an and bn is absolutely convergent, and n=0 n=0 +∞ +∞ ! +∞ ! X X X cn = an bn . n=0 n=0 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 cn = ajbk and an bn = ajbk. Also n=0 (j,k)∈Tm n=0 n=0 (j,k)∈Sm m +∞ ! +∞ ! X X X X X |cn| ≤ |aj||bk| ≤ |aj||bk| ≤ |an| |bn| , n=0 (j,k)∈Tm (j,k)∈Sm n=0 n=0 n +∞ +∞ P P P since |cn| ≤ |aj||bn−j| and the infinite series an and bn are absolutely j=0 n=0 n=0 +∞ P convergent. It follows from Proposition 2.2 that the Cauchy product cn n=0 is absolutely convergent, and is thus convergent. Moreover 2m m ! m ! X X X cn − an bn n=0 n=0 n=0 X = ajbk (j,k)∈T2m\Sm X X ≤ |ajbk| ≤ |ajbk| (j,k)∈T2m\Sm (j,k)∈S2m\Sm 2m ! 2m ! m ! m ! X X X X = |an| |bn| − |an| |bn| , n=0 n=0 n=0 n=0 since Sm ⊂ T2m ⊂ S2m.
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