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. Monday, March 24 9.4 Other tests p473: 2,14,26,39 Main Idea: Bigger than big is _________ Smaller than small is ___________ Bigger than small is _____________ Smaller than big is ____________________ Key Words: (a) Comparison Test, (b) Limit Comparison Test, (c) Ratio test Goal: Notice how I write out the proofs and then do likewise. -------------------------------------------------------- Previous assignment: p467: (16),(18),(22),(26) Page 467 Problem 16 Infinity 1 1 Infinity 1 Infinity 1 SUM ---- + ------ = SUM ---- + SUM ------ k=1 2 k k=1 2 k=1 k k 2 k 2 Infinity 1 SUM ----- is a p-series, p = 2 so it converges. k=1 2 k Infinity 1 SUM ----- is a geometric series, r = 1/2 so it converges. k=1 k 2 ---------------------------------------------------------------- Get["font.math"]; f[n_] := Sum[ (1/k^2 + 1/2^k),{k,1,n}]; A = Table[f[n],{n,1,50}]; a = ListPlot[A,PlotRange->All,PlotLabel->"P467 p16; SUM 1/k^2+1/2^k"]; b = Plot[Pi^2/6 + 1,{x,0,50},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotRange->All, PlotLabel->"P467 p16: SUM 1/k^2 + 1/2^k"]; Display["p16x.ps",c]; p[n_] := Sum[ (1/k^2 ),{k,1,n}]; BB = Table[p[n],{n,1,50}]; bb = ListPlot[BB,PlotRange->All,PlotStyle->{RGBColor[1,0,0]}]; cc = Plot[Pi^2/6 ,{x,0,50},PlotStyle->{RGBColor[0,1,0]}]; g[n_] := Sum[ (1/2^k ),{k,1,n}]; BBB = Table[g[n],{n,1,50}]; bbb = ListPlot[BBB,PlotRange->All,PlotStyle->{RGBColor[1,0,0]}]; ccc = Plot[1 ,{x,0,50},PlotStyle->{RGBColor[0,1,0]}]; z = Plot[0,{x,0,50}]; ddd = Show[c,bb,cc,bbb,ccc,z,PlotRange->All, PlotLabel->"P467 p16: SUM 1/k^2 + 1/2^k"]; Display["p16y.ps",c]; ---------------------------------------------------------------- Page 467 Problem 18 _ _ Infinity | 1 | SUM k Sin| ---- | k=1 |_ k _| _ _ | 1 | Limit k Sin| ---- | = 1 k->Infinity |_ k _| Therefore the series diverges by the n-th term test. ---------------------------------------------------------------- Get["font.math"]; g[n_] := n Sin[1/n]; B = Table[g[n],{n,1,50}]; b = ListPlot[B,PlotRange->All,PlotLabel->"P467 p18 y = n Sin[1/n]"]; c = Plot[0,{x,0,50}]; d = Show[b,c,PlotLabel->"P467 p18 y = n Sin[1/n]"]; Display["p18x.ps",d]; f[n_] := Sum[ k Sin[1/k],{k,1,n}]; A = Table[f[n],{n,1,500}]; a = ListPlot[A,PlotRange->All,PlotLabel->"P467 p18; SUM k Sin[1/k] "]; Display["p18y.ps",a]; ---------------------------------------------------------------- Page 467 Problem 22 Infinity 1 SUM ---------- k=1 2 1+4 k ........................................................ Infinity 1 Infinity 1 SUM ---------- < SUM ------- k=1 2 k=1 2 1+4 k 4 k This is a p series, p = 2 and converges. Thus the original series converges by the comparison test. ........................................................ 1 (a) f(x) = --------- is positive and decreasing. 2 1+4x Infinity 1 | x=Infinity (b) INT -------------- = 1/2 ArcTan[2x] | x=1 2 | x=1 1 + (2x) = 1/2( Pi/2 - ArcTan[2]) which converges. (c) Therefore the original series converges by the integral test. ---------------------------------------------------------------- Get["font.math"]; f[n_] := Sum[1.0/(1+4 k^2),{k,1,n}]; g[n_] := Sum[1/(4 k^2),{k,1,n}]; A = Table[f[n],{n,1,50}]; B = Table[g[n],{n,1,50}]; P1 = ListPlot[A]; P2 = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; P3 = Plot[Pi^2/24,{x,0,50},PlotStyle->{RGBColor[0,0,1]}]; P4 = Show[P1,P2,P3, PlotLabel->"P467 p22,black=SUM 1/(1+4k^2); Red is p-series p = 2"]; Display["p22x.ps",P4]; P5 = Plot[ 1/2(Pi/2 - ArcTan[2]),{x,0,50},PlotStyle->{RGBColor[0,0,1]}]; AA = Table[f[n]-f[1],{n,1,50}]; P6 = ListPlot[AA]; P7 = Show[P1,P5,P6,PlotLabel->"P467 p22 Sum 1/(1+4k^2) and Integral and no first term"]; Display["p22y.ps",P7]; ---------------------------------------------------------------- Page 467 Problem 26 Estimate the error that is made by approximating the sum of the given series by the sum of the first five terms. We first calculate the actual limit and the value of the first five terms. We subtract them to get the exact error. Then we "bound" the error using the integral test. .......................................................... Computing the exact error. Infinity 1 Infinity 1 1 SUM ------------- = SUM ------ - ----- k=1 k(k+1) k=1 k k+1 k 1 2 3 4 5 1-1/2 1/2-1/3 1/3-1/4 1/4-1/5 1/5-1/6 The sum of the first 5 terms is 1-1/5 or 5/6. The sum of the whole series is 1. Thus the error is 1/6. ......................................................... Bounding the error using the integral test. 1 f(x) = 1/k - 1/(k+1) = ---------- is positive and decreasing k(k+1) _ _ Infinity 1 1 | |x=Infinity INT ----- - --- = | ln(x) - ln(x+1) | x=5 x x+1 |_ _|x=5 _ _ | | x=Infinity = | ln( x/(x+1)) | = 0 - ln(5/6) = 0.182322 |_ _| x=5 The bound on the error is greater than the error. 0.182322 > 0.166667 ---------------------------------------------------------------------- Get["font.math"]; f[n_] := Sum[1/(k(k+1)),{k,1,n}]; A = Table[f[n],{n,1,50}]; a = ListPlot[A]; b = Plot[1,{x,1,50},PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x,0,50}]; d = Show[a,b,c,PlotLabel->"P467 p26 SUM 1/(k(k+1))",PlotRange->All ]; Display["p26.ps",d]; ---------------------------------------------------------------------- Old Material: n-th term test. If Limit a does not exist, the series diverges. n->Infinity n If Limit a exists and is not zero, the series diverges. n->Infinity n harmonic series: 1 + 1/2 + 1/3 + ... + 1/n + ... diverges 2 3 n geometric series: a + ar + ar + ar + ... + ar + ... diverges if |r| >=1, first term a converges to ------------ = ---------- if |r| < 1. 1-ratio 1-r 1 1 1 p-series: 1 + ------- + ------- + ... + ------- + ... p p p 2 3 n converges if p > 1 diverges if 0 <= p <= 1. Integral test: For f(x) positive and eventually decreasing Infinity f(1) + f(2) + f(3) + ... converges <=> INT f(x) dx converges x=1 New Material; Page 448 THEOREM A. Comparison test: If 0 <= a <= b for n >= N. (* eventually *) n n then (i) If SUM a diverges then SUM b diverges. n n (ii) If Sum b converges then SUM a converges. n n n Page 469 Example 1. Does SUM ----------- converge or diverge. 2 5 n - 4 (* A simple YES or NO will not suffice. You must use a test *) n n 1 (a) SUM ------- >= SUM ------ = (1/5) SUM ----- 2 2 n 5n -4 5n (b) This is the harmonic series which diverges. (c) Thus the original series diverges by the Comparison Test. Page 469 Example 2. n Does SUM ---------- Converge or diverge. n 2 (n+1) n 1 (a) SUM --------- <= SUM ----- n n 2 (n+1) 2 (b) This is a geometric series, r = 1/2 which converges. (c) Therefore the original series converges by the Comparison Test. LIMIT COMPARISON TEST Page 470 THEOREM B. Suppose that a >= 0 and b > 0 n n a n and Lim ----- = L b n If 0 < L < Infinity, then SUM a and SUM b both converge or both diverge. n n If L=0, then if SUM b converges, then also SUM a Converges. n n Page 470 Example 3: Determine the convergence of divergence of each series. Infinity 3n-2 (a) SUM --------------- n=1 3 2 n - 2n + 11 The trick is picking a suitable b . Simply compute b from the n n highest powers of n in the numerator and the denominator. n 1 Let b = ----- = -----. n 3 2 n n 2 3 2 3n-2 n 3n - 2n Limit ---------------- * ---- = Limit --------------- = 3 3 2 1 3 2 n - 2n + 11 n - 2n + 11 SUM b is a p series, p = 2 and converges. n Therefore the original series converges by the limit comparison test. ........................................................................ Infinity 1 (b) SUM --------------------- n=1 _ _ | 2 | Sqrt| n + 10 n | |_ _| 1 Let b = --- n n 1 n n 1 Limit ------------ * --- = Limit ------------ = Limit ------------- = 1 _ _ 1 _ _ _ _ | 2 | | 2 | | 10 | Sqrt| n +10 n| Sqrt| n + 10 n| Sqrt| 1 + --- | |_ _| |_ _| |_ n _| SUM b is the harmonic series and diverges. n Therefore the original series diverges by the limit comparison test. Page 471 RATIO test. Let SUM a be a series of positive terms and suppose that n a n+1 Limit ------ = p a n (a) If p < 1 the series converges. (b) If p > 1 the series diverges. (c) If p = 1 the test is inconclusive. Page 472 Example 5. Test for convergence or divergence n Infinity 2 SUM -------- n=1 n! a n+1 n+1 2 n! 2 ------ = ----- * ----- = ------------ = 0 a (n+1)! n n+1 n 2 By the ratio test r = 0, the series converges. ------------------------------------------------------ Get["font.math"]; f[n_] := 2^n/n!; A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,20}]; a = ListPlot[A]; b = Plot[E^2 - 1,{x,1,20},PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x,0,20}]; d = Show[a,b,c,PlotLabel->"P451 ex5; SUM 2^n/n!"]; Display["ex5.ps",d]; -------------------------------------------------------------- Page 472 Example 6 n Infinity 2 SUM -------- n=1 20 n a n+1 20 _ _ 20 n+1 2 n | n | ------ = ------- * ------- = 2 | ------ | = 2 a 20 n |_ n+1 _| n (n+1) 2 By the ratio test r = 2, the series diverges. ------------------------------------------------------ Get["font.math"]; f[n_] := 2^n/n^20; A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,100}]; a = ListPlot[A]; b = Plot[0,{x,1,100},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P451 ex6; SUM 2^n/n^20",PlotRange->All]; Display["ex6x.ps",c]; f[n_] := 2^n/n^20; A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,150}]; a = ListPlot[A]; b = Plot[0,{x,1,150},PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x,0,150}]; d = Show[a,b,c,PlotLabel->"P451 ex6; SUM 2^n/n^20"]; Display["ex6y.ps",d]; ------------------------------------------------------ Page 472 Example 7 Infinity n! SUM ------------ n=1 n n a n n n+1 (n+1)! n (n+1) n ------- = ------------ * ------ = ----------- a n+1 n! n+1 n (n+1) (n+1) _ _ n | n | = | ----- | which approaches 1/e < 1. |_ n+1 _| By the ratio test r=1/e, the series converges. --------------------------------------------------- Get["font.math"]; f[n_] := n!/n^n; A = Table[{n,Sum[f[i],{i,1,n}]},{n,1,50}]; a = ListPlot[A]; b = Plot[0,{x,1,50},PlotStyle->{RGBColor[1,0,0]}]; c = Plot[0,{x,0,50}]; d = Show[a,b,c,PlotLabel->"P451 ex6; SUM n!/n^n"]; Display["ex7.ps",d]; ------------------------------------------------