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.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. March 11 10.5 Alt Series p457: 1-3,(4),(14),15-17,21,(28),29,(40) ######################################################### # # # # # # # NO CLASS FOR A WHOLE WEEK...Yahoo! # # # # # # # # # # # ######################################################### Main Idea: + - + - + - + - ... and decreasing to zero are wonderful things. Key Words: Alternating Series, Absolute Convergence, Conditionally Convergent, First Excluded Term Goal: Learn techniques for alternating series and also techniques for absolute convergence. -------------------------------------------------------- Previous assignment: Page 452 Problem 4 Use the Limit Comparison Test to determine convergence or divergence. Infinity Sqrt[2n+1] SUM ------------------ n=1 2 n Sqrt[n] 3/2 Compare with ------- = n 2 n _______ 2 / 2n+1 Sqrt[2n+1] n / ------ = Sqrt[2] Limit -------------- * ------------ = Limit \ / n n->Infinity 2 n->Infinity \/ n Sqrt[n] Sqrt[n] 3/2 Since -------- = (1/n) is a p series, p=3/2 > 1 it converges. 2 n Therefore the original series converges by the limit comparison test. -------------------------------------------------------------------- Page 452 Problem 17 3 Infinity 4 n + 3 n SUM ----------------- n=1 5 2 n - 4 n + 1 3 n 1 Compare with -----= ----. 5 2 n n 3 4 + ----- 3 5 2 4 n + 3 n n n --------------- * ----- = ------------------ = 4 5 2 3 -4 1 n - 4n + 1 n 1 + ----- + ----- 3 5 n n 3 n 1 Since ------ = ---- is a p series, p=2 > 1 it converges. 5 2 n n Therefore the original series converges by the limit comparison test. --------------------------------------------------------------------- Page 452 Problem 25 1 1 1 1 1 + ------- + ----------- + ----------- + ---------- + ... 2 Sqrt[2] 3 Sqrt[3] 4 Sqrt[4] 5 Sqrt[5] This is a p-series, p = 3/2 > 1 so it converges. ------------------------------------------------------------- Page 452 Problem 43 Test for convergence or divergence. In some cases, a clever manipulation using the properties of logarithms will simplify the problem. Infinity (a) SUM ln( 1 + 1/n) n=1 | 2 | Infinity | (n+1) | (b) SUM ln | -------- | n=1 | n(n+2) | Infinity 1 (c) SUM -------- n=2 ln(n) ln(n) Infinity 1 (d) SUM ---------- n=2 ln(n) ln(ln(n)) Infinity 1 (e) SUM ---------------- n=2 4 ln(n) _ _ 2 Infinity | ln(n) | (f) SUM | ----- | n=1 |_ n _| --------------------------------------------- Infinity (a) SUM ln( 1 + 1/n) n=1 k SUM ln( n + 1) - ln(n) = ln(k+1) n=1 It is a telescoping series and diverges ----------------------------------------- f[x_] := Log[1+1/x]; A = Table[ {i,Sum[f[j],{j,1,i}]},{i,1,20}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; b = Plot[Log[x+1],{x,0,20}]; y = ParametricPlot[{0,y},{y,0,2}]; c = Show[a,b,y,PlotLabel->"P452 P43(a) red = SUM ln(x+1/x) black=Log[x+1]"]; Display["43a.ps",c]; -------------------------------------------- | 2 | Infinity | (n+1) | (b) SUM ln | -------- | n=1 | n(n+2) | f[x] = 2 ln(x+1) - ln(x) - ln(x+2) -2 f'(x) = 2/(x+1) - 1/x -1/(x+2) = ----------------- x(x+1)(x+2) So f[x] is decreasing. a | | x=a INT f[x] dx = | -(x Log[x]) + 2 (1 + x) Log[1 + x] - (2 + x) Log[2 + x] | 1 | | x=1 = -a ln(a) + 2 (1+a) ln(1+a) -(2+a) ln(2+a) - 4 ln(2) + 3 ln(3) 2(1+a) (1+a) = ln -------------------- - 4 ln(2) + 3 ln(3) a 2+a a (2+a) (1+a) (1+a) (1+a) (1+a) = ln -------------------- - 4 ln(2) + 3 ln(3) a 2+a a (2+a) a (2+a) (1+a) (1+a) = ln -------------------- - 4 ln(2) + 3 ln(3) a 2+a a (2+a) | |a | |(2+a) | 1+a| | 1+a | = ln |----| |-----| - 4 ln(2) + 3 ln(3) | a | | 2+a | | |a | |(2+a) | 1+a| | 1+a | = ln |----| + ln |-----| - 4 ln(2) + 3 ln(3) | a | | 2+a | | |a | |(2+a) | 1| | 2+a-1 | = ln |1+ -| + ln |----- | - 4 ln(2) + 3 ln(3) | a| | 2+a | | |a | |(2+a) | 1| | 1 | = ln |1+ -| + ln |1 - --- | - 4 ln(2) + 3 ln(3) | a| | 2+a | ln (e) + ln(1/e) - 4 ln(2) + 3 ln(3). Therefore by the integral test, the original series converges. ----------------------------------- f[x_] := Log[(x+1)^2 / (x(x+2)) ]; g[x_] := -(x Log[x]) + 2 (1 + x) Log[1 + x] - (2 + x) Log[2 + x]; A = Table[ {i,Sum[f[j],{j,1,i}]},{i,1,200}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; b = Plot[ g[x]-g[1], {x,0,200}]; y = ParametricPlot[{0,y},{y,0,2}]; c = Plot[-4 Log[2]+3 Log[3],{x,0,200},PlotStyle->{RGBColor[0,0,1]}]; d = Show[a,b,c,y, PlotLabel->"P452 P43(b) red = SUM ln (x+1)^2/(x(x+1)) and black=INT-> -4ln(2)+3ln(3) ", PlotRange->All]; Display["43b.ps",d]; -------------------------------------------- Infinity 1 Infinity 1 (c) SUM --------------- = SUM _------------_ n=2 ln(n) n=2 | ln(n) | ln(n) ln|_ ln(n) _| e Infinity 1 Infinity 1 = SUM ---------------- = SUM ------- n=2 ln(n) ln(ln(n)) n=2 ln(ln(n)) e n Infinity 1 <= SUM --------- eventually. n=2 2 n This is a p series, p = 2 and converges. The original series converges by the comparison test. ----------------------------------------------------------------------- f[x_] := 1/ Log[x]^Log[x]; g[x_] := 1/x^2; a = Plot[f[x],{x,1000,2000},PlotStyle->{RGBColor[1,0,0]}]; b = Plot[g[x],{x,1000,2000},PlotStyle->{RGBColor[0,0,1]}]; c = ParametricPlot[{E^(E^2),y},{y,0,f[E^(E^2)]},PlotStyle->{RGBColor[0,1,0]}]; d = Show[a,b,c,PlotLabel->"P452 P43(c) red = 1/ln(x)^ln(x) blue = 1/n^2"]; Display["43c.ps",d]; ---------------------------------------------------------- Infinity 1 (d) SUM ---------- n=2 ln(n) ln(ln(n)) Infinity 1 = SUM ---------- n=2 ln(n) ln(ln(ln(n))) e Infinity 1 = SUM ------------------- n=2 ln(ln(ln(n))) n Infinity 1 702 < = SUM --------------- when n > E^(E^(E^2)) or 5 10 n=2 2 n This is a p series, p=2 which converges. The original series converges by the comparison test. ---------------------------------------------------------- g[x_] := 1/x^2; h[x_] := 1/Log[Log[x]]^Log[x]; a = Plot[Pi^2/6,{x,3,400},PlotStyle->{RGBColor[1,0,0]}]; B = Table[ {i,Sum[g[j],{j,1,i}]},{i,1,400}]; b = ListPlot[B,PlotStyle->{RGBColor[0,0,1]}]; U = Table[ {i,Sum[h[j],{j,3,i}]},{i,3,400}]; u = ListPlot[U,PlotStyle->{RGBColor[0,1,0]}]; y = ParametricPlot[{0,y},{y,0,2}]; d = Show[b,y,u,a, PlotLabel->"P452 P43(d) red=Pi^2/6 blue=SUM 1/n^2 green 1/ln(ln(n))^ln(n) ", PlotRange->All]; Display["43d.ps",d]; ----------------------------------------------------------------- Infinity 1 (e) SUM ---------------- n=2 4 ln(n) 4 ln(n) 4 ln(n)^3 (1/n) Limit -------- = Limit ------------ n->Infinity n n->Infinity 1 4 ln(n)^3 12 ln(n)^2 1/n Limit -------------- = Limit ------------ n->Infinity n n->Infinity 1 12 ln(n)^2 24 ln(n) 1/n Limit -------------- = Limit ------------ n->Infinity n n->Infinity 1 24 ln(n) 24 1/n Limit -------------- = Limit ------------ n->Infinity n n->Infinity 1 24 Limit ---- = 0. Thus eventually n->Infinity n ln(n)^4 < n for n > No and then 1 -------- > 1/n for n > No ln(n)^4 Infinity 1 Infinity SUM ---------------- > SUM 1/n No 4 No ln(n) This is the harmonic series which diverges. Therefore the original series also diverges by the comparison test. --------------------------------------------------------------- f[x_] := 1/ Log[x]^4; g[x_] := 1/x; A = Table[ {i,Sum[f[j],{j,5,i}]},{i,2,1000}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; B = Table[ {i,Sum[g[j],{j,5,i}]},{i,2,1000}]; b = ListPlot[B,PlotStyle->{RGBColor[0,0,1]}]; y = ParametricPlot[{0,y},{y,0,2}]; d = Show[a,b,y, PlotLabel->"P452 P43(e) red=SUM 1/ln(n)^4 blue=SUM 1/n ", PlotRange->All]; Display["43e.ps",d]; --------------------------------------------------------- f[x_] := 1/ Log[x]^4; g[x_] := 1/x; p = Plot[f[x],{x,4000,6000}]; q = Plot[g[x],{x,4000,6000},PlotStyle->{RGBColor[1,0,0]}]; r = Show[p,q,PlotLabel->"P452 P43(e) RED SHOULD BE BELOW"]; Display["43ex.ps",r]; -------------------------------------------------------- _ _ 2 Infinity | ln(n) | (f) SUM | ----- | n=1 |_ n _| _ _ 2 Infinity | ln(n) | Infinity 1 SUM | ----- | > SUM ------- eventually. n=1 |_ n _| n=1 3/2 n 2 | ln(n) | | -------| | n | ln(n)^2 Limit ---------- = Limit ---------- = 0 n->Infinity 1 n->Infinity n^(1/2) ------- 3/2 n Therefore from some No onwards 2 | ln(n) | 1 | ------ | < ------- | n | 3/2 n This is a p-series, p = 3/2 and converges. Therefore the original series converges by the comparison test. -------------------------------------------------------------- r = 6000; s = 6200; f[x_] := (Log[x]/x)^2; g[x_] := 1/x^(3/2); A = Table[ {i,Sum[f[j],{j,5,i}]},{i,r,s}]; a = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; B = Table[ {i,Sum[g[j],{j,5,i}]},{i,r,s}]; b = ListPlot[B,PlotStyle->{RGBColor[0,0,1]}]; y = ParametricPlot[{0,y},{y,0,2}]; d = Show[a,b,y, PlotLabel->"P452 P43(f) red=SUM (ln(n)/n)^2 < blue=SUM 1/n^(3/2)", PlotRange->All]; Display["43f.ps",d]; ---------------------------------------------------------------------------- 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 +... ..................................................... 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. But on the other hand... 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. 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. ===================================================================== ALTERNATING SERIES TEST (a) If {an} decreases to zero then a1-a2+a3-a4+a5 ... converges. (b) If the series converges to S and Sn is the sum of the first n terms, then |Sn-S| <= an+1. (* The error is less than the first omitted term *) Proof: Same as for the alternating series: 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 = a1-a2+a3-a4+a5-a6+a7-a8+a9 ... +a - a 2n 2n-1 2n S = (a1-a2)+(a3-a4)+(a5-a6)+(a7-a8)+(a9-a10) ... +(a -a ) 2n 2n-1 2n So S > 0 2n S = a1-(a2-a3)-(a4-a5)-(a6-a7)-(a8-a9) ... -(a - a ) - a 2n 2n-2 2n-1 2n So S is positive and bounded by a1. 2n 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 <= a1. In other words, the absolute value of the sum is no larger than a1. 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 |an| converges, then SUM an converges. Proof: 0 <= an + |an| <= 2 |an| By the comparison test SUM an + |an| converges. Therefore Sum an = SUM an+|an| - SUM |an| converges as well. 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. But if we have a non-positive term series, then we can 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. Example 4. Show that n Infinity n+1 3 SUM (-1) ---------- i=1 n! Converges absolutely. n+1 3 n! 3 (a) |an+1|/|an| = ------- * ------ = ---- which approaches 0 (n+1)! n n+1 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!] Example 5. Show that SUM -------- converges. 2 n n+1 1 Example 6 Show that SUM (-1) -------- is conditionally convergent. Sqrt[n]