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MAT 5610, Analysis I Part III

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MAT 5610, Analysis I Part III Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series MAT 5610, Analysis I Part III Wm C Bauldry [email protected] Spring, 2011 MAT 5610: 0 0 Today Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series MAT 5610: 0 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Outerlude MAT 5610: 19 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Seriesly Fun Definition • A series is a sequence of partial sums with the nth term given by Pn Sn = ak for some fixed p Z k=p 2 Pn • A sequence bn n p can be written as a series Sn = k p ak by f g ≥ = setting an = bn bn 1 and bp 1 = 0: − − − Examples X1 1 π2 X1 ( 1)k+1 1. = 4. − = ln(2) k2 6 k k=1 k=1 X1 1 k 2. = e X1 ( 1) +1 sin(k) k! 5. − = k=0 k 1 1 k=1 X 1 X 1 3. =1= sin(1) k(k + 1) k(k − 1) arctan k=1 k=2 1 + cos(1) MAT 5610: 20 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Harmonious Series RECALL: Definition P A series 1 ak converges to A R iff for every " > 0 there is an k=p 2 n∗ N s.t. whenever n n∗, then Sn A < ": 2 ≥ j − j Proposition P • If ak converges, then ak 0: P ! • If ak 0; then ak diverges. 6! Example (Nicole Oresme (c.1360)) The harmonic series P 1=k diverges. Take segments that are greater than 1=2 to show S n 1 + n=2. 2 ≥ MAT 5610: 21 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Critical Series Proposition The geometric series P a rk converges iff the constant r < 1, then j j X1 a a rk = : 1 r k=0 − Proof. 1. If r 1; then a rk 0. Whence P a rk diverges. j j ≥ 6! n+1 2. Let r < 1 and n > 1. Then Sn rSn = a a r . Hence j j − − a(1 rn+1) Sn = − 1 r − n 3. Since r < 1; then r 0: Thence Sn converges. j j ! What about P sin(k)k? MAT 5610: 22 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Geometric Sample Example Write 0:1011 as a rational number. 4 42 43 1. 0:1011 = 1011 10− + 1011 10− + 1011 10− + · · · ··· X1 4k 2. = 1011 10− · k=1 ! X1 4k 3. = 1011 1 + 10− − k=0 1 4. = 1011 1 + − 1 10 4 − − 10000 1 337 5. = 1011 1 + = 1011 = − 9999 9999 3333 MAT 5610: 23 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Maple Sample Maple Summing X1 ( 1)k+1 sin(k) sin(1) How does Maple find that − = arctan ? k 1 + cos(1) k=1 1. First, enter > infolevel[sum]:=5: to raise the level of “user information” printed. 2. Now enter the summation X1 ( 1)k+1 > − sin(k) k k=1 and wait for the results.1 1PlanB. MAT 5610: 24 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Cauchy and Friends Theorem (Cauchy Criterion for Series) P ak converges iff for each " > 0 there is an n∗ N s.t. whenever 2 n m n∗; then ≥ ≥ Sn Sm 1 = am + am+1 + am+2 + + an < " j − − j j ··· j Proof. X (Series are sequences; cf. Cauchy’s Criterion for Sequences.) Corollary P P 1. If ak = bk for k k∗, and bk converges, then ak converges. P ≥ P 2. If ak converges, then ak converges. P j j P 3. If ak diverges, then ak diverges j j MAT 5610: 25 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, I Theorem (The Integral Test) If f :[1; ) R is continuous, decreasing, and nonnegative, then 1 ! Z X1 1 f(k) converges iff f(x) dx converges k=1 1 Proof. Let ak = f(k): Since f is decr and cont, ak f(x) ak for k 1: +1 ≤ ≤ ≥ Whence Z k+1 ak+1 f(x) dx ak ≤ k ≤ Therefore Z k+1 Sk+1 a1 f(x) dx Sk − ≤ 1 ≤ The series bounds the integral, the integral bounds the series. Since all terms are nonneg, Bolzano-Weierstrass Thm gives the result. MAT 5610: 26 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Integral Test Figure 1 1 X Z 1 Z 1 X ak ≤ f(x) dx f(x) dx ≤ ak k=2 1 1 k=1 Z Z 1 X1 1 X1 So f(x) dx ak f(x) dx ak ≤ ≤ ≤ 2 k=2 1 k=1 MAT 5610: 27 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, II Theorem (The Comparison Test) P P Suppose ak and bk are two series with bk 0 for all k. ≥ + P 1. If ak Mbk eventually (for some M R ) and bk converges, j jP ≤ P 2 then ak and ak both converge. j j + P 2. If mbk ak eventually (for some m R ) and bk diverges, P≤ j j 1 2 then ak diverges. j j Proof. P 1. Wolog ak Mbk for all k. Since bk converges, so does P j j ≤ Mbk, say to S. Then N N X X X1 0 ak Mbk Mbk = S ≤ j j ≤ ≤ k=1 k=1 k=1 P Whence ak is a bounded, increasing series, thus convergent. j j 1 P k+1 ak may or may not diverge; e.g., ak = (−1) =k and bk = 1=k. MAT 5610: 28 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series A Useful Series Set Proposition (p-Series1) 1 X 1 The real series ζ(p) = converges iff p > 1: kp k=1 Proof. X (Apply the integral test.) Examples 1 1 X X 4 4 1. 1=k = 1 4. 1=k = π =90 k=1 k=1 1 1 X 2 2 X 5 2. 1=k = π =6 5. 1=k = ζ(5) k=1 k=1 1 1 X 3 X 6 6 3. 1=k = ζ(3) 6. 1=k = π =945 k=1 k=1 1See the Riemann Zeta function. Also $ MAT 5610: 29 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, III Theorem (The Limit Comparison Test) P P Suppose ak and bk are two positive-term series. an P P 1. If lim is finite and bn converges, then an converges. n!1 bn bn P P 2. If lim is finite and bn diverges, then an diverges. n!1 an Proof. Set L = lim an=bn. + 1. If L = 0; then 9M 2 R s.t. an ≤ Mbn. Now compare. 1 3 2. If L > 0; then 2 L bn ≤ an ≤ 2 L bn eventually. Again compare. Corollary P P an Suppose ak & bk are positive-term series, and lim = L with n!1 b P P n 0 < L < 1: Then ak & bk converge or diverge together. MAT 5610: 30 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Fun Time Exercises 1 1 p 1 X (−1)kk X ( 5 − 1)m X (−1)nn 1. 5. 9. (k + 1) ek m2 + 1 n2 + 1 k=1 m=1 i=1 1 1 1 X 102n X k! X (−1)mm3 2. 6. 10. (2n − 1)! kk (m2 + 1)4=3 n=1 k=1 m=1 1 j 1 n−1 1 1·4 1·4·7 X 3 X (−1) 11. 3 + 3·6 + 3·6·9 +··· 3. 7. j3 n2 + 1 2 2·5 2·5·8 j=1 n=1 12. 9 + 9·12 + 9·12·15 + 1 1 j 1 X (−1)i 23i X (−1) X k + 2 4. 8. 13. p 32i j ln(j) i=1 j=2 k=1 (k+1) k+3 1 X k a − (a − b)r Challenge: (a + kb)r = ; jrj < 1 (1 − r)2 k=1 (cacdcadcac2c2cddcd) MAT 5610: 31 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, IV Theorem (Ratio Comparison Test) P P Let an & bn be positive term series. If an+1=an ≤ bn+1=bn eventually, P P and bn converges, then an converges. Theorem (D’Alembert’s Ratio Test) P Let an be a series with positive terms. If there is a constant α 2 [0; 1) s.t. P an+1=an ≤ α eventually, then an converges. If an+1=an ≥ 1 eventually, P then an diverges. Corollary (Cauchy’s Ratio Test) P an+1 Let an be a series with positive terms. Set r = limn!1 : Then: an • If r < 1; the series converges. • If r > 1; the series diverges. • If r = 1; the test fails. MAT 5610: 32 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Proofs Proof (Ratio Comparison Test). 1. an=bn is decreasing eventually, thus bounded. Thus an Mbn. P P ≤ 2. Since bn converges, then an converges by comparison. Proof (D’Alembert’s Ratio Test). 1. α < 1 gives P αn converges. n+1 n P 2. Then an =an α /α = α implies that an converges. +1 ≤ Proof (Cauchy’s Ratio Test). X MAT 5610: 33 Introduction Review Derivative Integral Riemann-Stieltjes Integral Sequences & Series Convergence Tests, V Theorem (Root Test) p Let P a be a series with positive terms. If 9α 2 [0; 1) s.t. n a ≤ α n p n P n P eventually, then an converges. If an > 1 eventually, then an diverges. Theorem (Cauchy’s Root Test) p P n Let an be a series with positive terms.
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