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. Friday, March 28 9.5 Conditional Convergence p505: 1-28 Give Reasons ----------------------------------------------------------- Previous assignment: page 478: 6,12,26,29,45 page 478 Problem 6 Show that each alternating series converges, and then estimate the error made by using the partial sum S as an approximation to the sum S of the series. 9 INF n+1 Ln[n] SUM (-1) ---------- n=1 Sqrt[n] th We first must show that the limit of the n term is zero. Ln[n] 1/n 1 Limit -------- = Limit --------- = Limit ---------- = 0 n->Inf Sqrt[n] n->Inf -1/2 n->Inf 2 Sqrt[n] 1/2 n We next must show that it is decreasing. Ln{x] f[x] = ---------- Sqrt[x] -1/2 / Sqrt[x] 1/x - Ln[x] 1/2 x f [x] = ----------------------------- x / 1 -1/2 Ln[x] f [x] = ----------------------------- 3/2 x / 1 - Ln[ Sqrt[x] ] f [x] = ------------------------ 3/2 x / 2 Since f [x] < 0 when x > e The sequence is eventually decreasing. Therefore the series converges by the alternating series test. The error is no more than the first omitted term Thus Sn has an error of at most Ln(n+1)/Sqrt[x+1]. S has an error of at most Ln[10]/Sqrt[10] = 0.728141 9 ----------------------------------------------------------- Get["font.math"]; f[n_] := Log[n]/Sqrt[n]; g[n_] := Sum[ (-1)^(i+1) f[i],{i,1,n}]; A = Table[g[n],{n,1,200}]; p1 = ListPlot[A]; p2 = Plot[g[9],{x,0,200},PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; p3 = Plot[ g[9]+f[10],{x,1,200}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; p4 = Plot[ g[9]-f[10],{x,1,200}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; p5 = Show[p4,p3,p2,p1,PlotLabel->"P478 P6 Sum (-1)^(n+1) Log[n]/Sqrt[n]", PlotRange->All]; Display["p6.ps",p3]; -------------------------------------------------------------------- page 478 Problem 12 Show that each series converges absolutely. n INF n+1 2 SUM (-1) ------- 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 sequence converges absolutely. --------------------------------------------------------- Get["font.math"]; f[n_] := (-1)^(n+1) 2^n/n!; g[n_] := Sum[f[i],{i,1,n}]; A = Table[g[n],{n,1,10}]; p1 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.008]}]; p2 = ParametricPlot[{0,y},{y,-1,1}]; p3 = Plot[0,{x,0,10}]; p4 = Show[p1,p2,p3,PlotLabel->"P478 p12 Sum (-1)^(n+1) 2^n/n!", PlotRange->All]; Display["p12.ps",p4]; ----------------------------------------------------------- page 478 Problem 26 Classify each series as absolutely convergent conditionally convergent, or divergent Inf Sum n Sin(1/n) n=1 Sin[1/n] Limit n Sin[1/n] = Limit --------- = 1 n->Inf n->Inf 1/n The sequence diverges by the n-th term test. -------------------------------------------------------- page 478 Problem 29 Classify each series as absolutely convergent conditionally convergent, or divergent n+1 Inf (-3) Sum ----- n=1 2 n | a | n+2 2 2 | n+1 | 3 n | n | Limit ---------- = Limit -------- ------ = Limit 3 |---| = 3 n->Inf | a | n->Inf 2 n+1 n->Inf |n+1| | n | (n+1) 3 The sequence of absolute values diverges by the ratio test, r=3. n+1 (-3) Limit a = Limit -------- The sequence Oscillates with n->Inf n n->Inf 2 n increasing amplitude so does not converge to 0. Therefore the series itself diverges by the n-th term test. ------------------------------------------------------------- page 478 Problem 45 Note that 1-1/2+1/2-1/4+ ... -1/2n = 1+1/2+1/3+ ... + 1/2n - (1+1/2+1/3+ ... + 1/n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n) Recognize the latter expression as a Riemann sum and use it to find the sum of the alternating harmonic series Partition [1,2] into n parts; The Riemann sum for f(x) = 1/x is n 1 n 1 Sum 1/n(--------) = Sum ------- i=1 (1+i/n) i=1 n+i Since the Riemann sum approaches the area under the curve y = 1/x on the interval [1,2], it converges to Log[2]. An alternate explanation: ---------------------------------------------------------------- The area under the curve y = 1/x from n to 2n is ln(2n)-ln(n) = ln(2) Thus: 1/(n+1) + 1/(n+2) + ... + 1/(2n) <= ln(2) <= (1/n) + 1/(n+1) + ....+1/(2n-1) S <= ln(2) < 1/n + S - 1/(2n) 2n 2n Thus the limit of S = ln(2). 2n ------------------------------------------------------------- Get["font.math"]; f[x_] := 1/x; g[n_] := Sum[(-1)^(i+1) f[i],{i,1,n}]; A = Table[g[n],{n,1,100}]; P1 = ListPlot[A]; P2 = ParametricPlot[{0,y},{y, 0, Log[2]}]; P3 = Plot[0,{x,0,100}]; P4 = Plot[Log[2],{x,0,100},PlotStyle->{RGBColor[1,0,0]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"P478 P45, Sum (-1)^(n+1) (1/n)",PlotRange->All ]; Display["p45x.ps",P5]; n = 10; B = Table[{0,0},{i,0,4(n+1)}]; Do[ B[[4i+1]] = {n+i, 0},{i,0,n}]; Do[ B[[4i+2]] = {n+i, 1/(n+i)},{i,0,n}]; Do[ B[[4i+3]] = {n+i+1, 1/(n+i)},{i,0,n}]; Do[ B[[4i+4]] = {n+i+1, 0},{i,0,n}]; P6 = ListPlot[B, PlotStyle->{RGBColor[1,0,0]},PlotJoined->True,PlotRange->{{0,2n},{0,0.1}}]; P7 = Plot[1/x,{x,n,2 n},PlotStyle->{Thickness[0.01]}]; P8 = Show[P6,P7,PlotLabel->"P478 P45 Area under 1/x from n to 2n"]; Display["p45y.ps",P8]; ////////////////////////////////////////////////////////////////////// Respond with true or false to each of the following assertions. Give an explanation to justify your answer. 1. If 0 <= a <= b for all natural numbers n n n and Limit b exists, then Limit a exists. n->Infinity n n->Infinity n ------------------------------------------------------------------ 2. For every positive integer n, n n! <= n <= (2n-1)! ------------------------------------------------------------------ 3. If Limit a = L, then Limit a = L n->Infinity n n->Infinity 3n+4 ------------------------------------------------------------------ 4. If Limit a = L and Limit a = L n->Infinity 2n n->Infinity 3n then Limit a = L n->Infinity n ------------------------------------------------------------------ 5. If Limit a = 0 for every positive integer m >= 2 n->Infinity mn then Limit a = L n->Infinity n ------------------------------------------------------------------ 6. If Limit a = L and Limit a = L n->Infinity 2n n->Infinity 2n+1 then Limit a = L n->Infinity n ------------------------------------------------------------------ 7. If Limit (a - a ) = 0, then Limit a exists and is finite. n->Inf n n-1 n->Inf n ------------------------------------------------------------------ 8. If {a } and {b } both diverge, then {a + b } diverges. n n n n ------------------------------------------------------------------ a n 9. If {a } converges, then {----} converges to 0. n n ------------------------------------------------------------------ INF INF 2 10. If SUM a converges, so does SUM a n=1 n n=1 n ------------------------------------------------------------------ 11. If 0 < a < a for all natural numbers n, and if n+1 n INF n+1 Limit a = 0, Then SUM (-1) a converges and has n->Infinity n n=1 n sum S satisfying 0 < S < a 1 ------------------------------------------------------------------ INF n 12. SUM (1/n) converges and has sum S satisfying 1 < S < 2. n=1 ------------------------------------------------------------------ 13. If a series SUM a diverges, then its sequence of partial sums is unbounded. n ------------------------------------------------------------------ INF 14. If 0 <= a <= b for all natural numbers n, and if SUM b diverges, then n n n=1 n INF SUM a diverges. n=1 n ------------------------------------------------------------------ 15. The Ratio Test will not help in determining the convergence or divergence of INF 2n+3 SUM ------------------------- n=1 4 3 3n + 2n + 3n + 1 ------------------------------------------------------------------ INF 16. If a > 0 for all natural numbers n and SUM a converges n n=1 n a n+1 then Limit ----- < 1 n->Infinity a n ------------------------------------------------------------------ _ _ n INF | 1 | 17. SUM | 1 - --- | converges. n=1 |_ n _| ------------------------------------------------------------------ INF 1 18. SUM ------------- converges. n=1 4 ln( n + 1) ------------------------------------------------------------------ INF n+1 19. SUM --------- converges. n=2 2 (n ln(n) ) ------------------------------------------------------------------ 2 n Pi Sin (------ ) INF 2 20. SUM ------------------ converges n=1 n ------------------------------------------------------------------ 21. If 0 <= a <= b for all natural numbers n n+100 n INF INF and if SUM b converges, then SUM a converges. n=1 n n=1 n ------------------------------------------------------------------ 22. If, for some c > 0, c a >= 1/n for all natural numbers n, n INF then SUM a diverges. n=1 n ------------------------------------------------------------------ 2 3 1000 23. (1/3) + (1/3) + (1/3) + ... + (1/3) < 1/2 TRUE: 1/3 The sum of this series is ------- = 1/2. So any partial sum is 1-1/3 less than 1/2. ------------------------------------------------------------------ INF INF n 24. If Sum a converges, then SUM (-1) a converges. n=1 n n=1 n ------------------------------------------------------------------ INF 25. If b <= a <= 0 for all natural numbers n, and if SUM b n n n=1 n INF converges, then SUM a converges. n=1 n ------------------------------------------------------------------ 26. If 0 <= a for all natural numbers n, and if n INF INF n SUM a converges, then SUM (-1) a converges. n=1 n n=1 n ------------------------------------------------------------------ | INF n+1 1 99 n+1 1 | 27. | SUM (-1) --- - SUM (-1) --- | < 0.01 | n=1 n n=1 n | ------------------------------------------------------------------ INF INF | | 28. If SUM a diverges then SUM | a | diverges. n=1 n n=1 | n |