Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Wednesday, March 26 9.5 Alt Series p478: 6,12,26,29,45 Main Idea: + - + - + - + - ... and decreasing to zero are wonderful things. Key Words: (a) Alternating Series, (b) Absolute Convergence, (c) Conditionally Convergent, (d) First Excluded Term Goal: Learn techniques for alternating series and also techniques for absolute convergence. -------------------------------------------------------- Previous assignment: p473: (2),(14),(26),(39) Page 473 Problem 2 Infinity 3 n + 1 SUM ------------- n=1 3 n - 4 3 n + 1 a = ------------ n 3 n - 4 n 1 (a) b = -------- = ---- n 3 2 n n a 2 n 3 n + 1 n (b) Limit ----- = Limit ------ * ------- n->Infinity b n->Infinity 3 1 n n - 4 3 2 3 n + n = Limit ---------------- = 3 n->Infinity 3 n - 4 (c) Sum b is a p-series, p=2 and converges. n (d) The original series converges by the limit comparison test. ---------------------------------------------------------------- Get["font.math"]; f[n_] := (3n+1)/(n^3-4); g[n_] := 1/n^2; A = Table[{n,f[n]/g[n]},{n,1,50}]; a = ListPlot[A]; b = Plot[0,{x,0,50}]; c = ParametricPlot[{0,y},{y,0,3}]; b = Show[a,b,c,PlotRange->All, PlotLabel->"P473 p2, an/bn=((3n+1)/(n^3-4))/(1/n^3)"]; Display["p2x.ps",b]; f[n_] := (3n+1)/(n^3-4); g[n_] := 1/n^2; A = Table[{n,f[n]},{n,30,50}]; B = Table[{n,g[n]},{n,30,50}]; a = ListPlot[A]; b = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x, 30,50}]; d = Show[a,b,c,PlotRange->All, PlotLabel->"P473 p2, terminal an and bn=red "]; Display["p2y.ps",d]; f[n_] := (3n+1)/(n^3-4); g[n_] := 1/n^2; A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,50}]; B = Table[{n,Sum[g[i],{i,1,n}]},{n,1,50}]; a = ListPlot[A]; b = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x, 1,50}]; d = Show[a,b,c,PlotRange->All, PlotLabel->"P473 p2, SUM an and SUM bn(red) "]; Display["p2z.ps",d]; ---------------------------------------------------------------- Page 473 Problem 14 Infinity Sqrt[n+1] SUM ------------ n=1 2 n + 1 Sqrt[n+1] a = ----------- n 2 n + 1 Sqrt[n] 1 (a) b = ----------- = ------- n 2 3/2 n n a 2 n Sqrt[n+1] n (b) Limit ----- = Limit --------- ---------- n->Infinity b n->Infinity 2 Sqrt[n] n n + 1 _ _ 2 | n+1 | n Limit Sqrt| ------ | * ------- = 1 n -> Infinity |_ n _| 2 n + 1 (c) Sum b is a p-series, p = 3/2 and converges. n (d) Therefore the original series converges by the limit comparison test. ---------------------------------------------------------------- Get["font.math"]; f[n_] := Sqrt[n+1]/(n^2+1); g[n_] := 1/n^(3/2); A = Table[f[n],{n,1,50}]; B = Table[g[n],{n,1,50}]; a = ListPlot[A,PlotStyle->{PointSize[0.02]}]; b = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotRange->All,PlotLabel->"P473 p14,black=Sqrt[n+1]/(n^2+1) "]; Display["p14.ps",c]; ---------------------------------------------------------------- Page 473 Problem 26 ln(2) ln(3) ln(4) ln(5) ------- + --------- + --------- + ------------ + ..... 2 2 2 2 2 3 4 5 ln(n) a = -------- n 2 n 1 (a) b = --------- n 3/2 n a 3/2 n ln(n) n (b) Limit ----- = Limit ----- * ----- n->Infinity b n->infinity 2 1 n n ln(n) 1/n = Limit -------- = Limit ------------ n->Infinity 1/2 n->Infinity -1/2 n 1/2 n 2 = Limit -------- = 0 n->Infinity Sqrt[n] (c) SUM b is a p-series, p = 3/2 and converges. n (d) The original series converges by the limit comparison test. ---------------------------------------------------------------- Get["font.math"]; f[n_] := Log[n]/(n^2); g[n_] := 1/n^(3/2); A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,50}]; B = Table[{n,Sum[g[i],{i,1,n}]},{n,1,50}]; a = ListPlot[A,PlotStyle->{PointSize[0.01]}]; b = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotRange->All, PlotLabel->"P473 p26,Partial Sums, black =ln(n)/n^2"]; Display["p26.ps",c]; ---------------------------------------------------------------- Page 473 Problem 39 Suppose that Limit n a = 1. Prove that SUM a diverges n->Infinity n n->Infinity n 1 (a) b = --- n n a n (b) Limit ----- = Limit n a = 1 n->Infinity b n->Infinity n n (c) SUM b is the harmonic series and diverges. n (d) Therefore the original series diverges by the limit comparison test. ================================================================ New Material x = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10 + 1/11 -1/12 +... 2x = 2 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6 + 2/7 - 2/8 + 2/9 - 2/10 + 2/11 -2/12 +... 2x = 2 - 1 + 2/3 - 1/2 + 2/5 - 1/3 + 2/7 - 1/4 + 2/9 - 1/5 + 2/11 -1/6 +... / < 2 + 2/3 + 2/5 + 2/7 + 2/9 + 2/11 +... \ 2x = > / < - 1 - 1/2 - 1/3 - 1/4 - 1/5 -1/6 +... \ / < 2 + 2/3 + 2/5 + 2/7 + 2/9 + 2/11 +... \ 2x = > / < -1 - 1/2 - 1/3 - 1/4 - 1/5 -1/6 - 1/7 - 1/8 - 1/9 - 1/10 - 1/11 + ... \ 2x = 1 -1/2 + 1/3 -1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 -1/10 + 1/11 + ... Therefore 2x = x so x = 1 or x = 0. x = (1-1/2) + (1/3-1/4) + (1/5-1/6) + (1/7-1/8) + (1/9-1/10) + (1/11-1/12) +... x = 1 - (1/2-1/3) - (1/4-1/5) - (1/6-1/7) - (1/8-1/9) - (1/10-1/11) - 1/12 +... The first says that x > 0 The second says that x < 1. What is going on? We showed that the sum is strictly between 0 and 1 and yet we also showed that it is either 0 or 1. Page 475 Theorem A: ALTERNATING SERIES TEST (a) If {a } decreases to zero then a -a +a -a +a ... converges. n 1 2 3 4 5 (b) If the series converges to S and S is the sum of the first n n terms, then |S -S| <= a . n n+1 (* The error is less than the first omitted term *) Proof: We study the partial sums with an even number of terms. We show that the even partial sums converge and then show that the odd partial sums converge to the same limit as the even partial sums do. S = a -a +a -a +a -a +a -a +a ... +a - a 2n 1 2 3 4 5 6 7 8 9 2n-1 2n S = (a -a )+(a -a )+(a -a )+(a -a )+(a -a ) ... +(a -a ) 2n 1 2 3 4 5 6 7 8 9 10 2n-1 2n So S > 0 2n S = a -(a -a )-(a -a )-(a -a )-(a -a ) ... -(a - a ) - a 2n 1 2 3 4 5 6 7 8 9 2n-2 2n-1 2n So S is positive and bounded by a . 2n 1 Thus the sequence S converges. 2n Suppose that S Converges to S. Then 2n Limit S = Limit S + (S - S ) = n->Infinity 2n+1 n->Infinity 2n 2n+1 2n Limit S + a = Limit S = S n->Infinity 2n 2n+1 n->Infinity 2n Since 0 <= S <= a , we have shown that 2n 1 0 <= S <= a . 1 In other words, the absolute value of the sum is no larger than a . 1 Applied to the tail of the series, the absolute value of the tail is no larger than the first term in the tail. Do these series converge? 1 1 1 1 1 1 1 1 - ------- + ------- - ------- + ------- - ------- + ------- - ------- ... Sqrt[2] Sqrt[3] Sqrt[4] Sqrt[5] Sqrt[6] Sqrt[7] Sqrt[8] 1 1 1 1 1 1 1 1 - ------- + ------- - ------- + ------- - ------- + ------- - ------- ... ln(2) ln(3) ln(4) ln(5) ln(6) ln(7) ln(8) Sin[1] - Sin[1/2] + Sin[1/3] - Sin[1/4] + Sin[1/5] - Sin[1/6] + Sin[1/7] ... 1 1 1 1 1 1 --- - --- + --- - --- + --- - --- + ... 1! 2! 3! 4! 5! 6! ================================================== When we have a series like Sin[1] + Sin[2] + Sin[3] + Sin[4] + ..... ----- ------ ----- ------ 1 2 3 4 It is not alternating, but it does have some negative signs. We attempt to determine whether it converges or diverges by dropping all the minus signs and testing the resulting all positive term sequence. Theorem If SUM |a | converges, then SUM a converges. n n Proof: Suppose that SUM | a | converges. Since n 0 <= a + | a | <= 2 |a | n n n _ _ | | By the comparison test SUM | a + |a | | converges. |_ n n _| _ _ | | Therefore Sum a = SUM | a + |a | | - SUM |a | converges as well. n |_ n n _| n If a series converges absolutely, then it converges, but there are series that converge but are not absolutely convergent. Such series are called CONDITIONALLY convergent and are some of the more interesting series. For Example 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ... Is conditionally convergent. For an absolutely convergent series, rearranging the terms is just fine. For a conditionally convergent series, rearranging the terms is a no-no. One can rearrange the terms to get the series to converge to any number you wish, even to converge to -Infinity or + Infinity. If we have a non-positive term series, then we often look at the absolute values and use the ratio test or the integral test. If we are lucky, we may be able to establish absolute convergence and then ordinary convergence follows. Page 477: Example 4. Show that n Infinity n+1 3 SUM (-1) ---------- i=1 n! Converges absolutely. |a | n+1 | n+1| 3 n! 3 (a) -----| = ------- * ------ = ---- which approaches 0 | a | (n+1)! n n+1 | n | 3 (b) The original series is absolutely convergent by the ratio test with r=0. (c) The original series converges by the absolute convergence test. Cos[n!] Page 477 Example 5: Show that SUM -------- converges. 2 n Page 478 Example 6 Show that n+1 1 SUM (-1) -------- is conditionally convergent. Sqrt[n]