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. Friday, March 28 9.5 Conditional Convergence p505: 1-28 Give Reasons 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 FALSE: This is a sequence: Counterexample: b = 1 for all n. n / 1 if n is even a = > n \ 0 if n is odd b converges to 1. a does not converge. n n 2. For every positive integer n, n n! <= n <= (2n-1)! TRUE: 1*2*3*4...*n < n*n*n*...*n both have same number of factors but those on the right are all bigger than or equal to the corresponding factors on the left. (2n-1)! = (2n-1)*(2n-2)* .... *(2n-n)*(n-1)! The first n terms on the right are all bigger or equal to n n Therefore n <= (2n-1)! 3. If Limit a = L, then Limit a = L n->Infinity n n->Infinity 3n+4 TRUE: If |a -L | < epsilon for all n > No, n then since 3n+4 > n, |a - L| < epsilon for all n > No 3n+4 4. If Limit a = L and Limit a = L n->Infinity 2n n->Infinity 3n then Limit a = L n->Infinity n FALSE: Let / 0 if n is not a prime a = > n \ 1 if n is a prime. a -------> 0 2n a ----------> 0 3n But Limit a does not exist. n 5. If Limit a = 0 for every positive integer m >= 2 n->Infinity mn then Limit a = L n->Infinity n FALSE: Use same counter example as was used in problem 4. 6. If Limit a = L and Limit a = L n->Infinity 2n n->Infinity 2n+1 then Limit a = L n->Infinity n TRUE: Given epsilon, There exists a N such that | a - L | < epsilon if n > N 1 2n 1 There exists a N such that | a - L | < epsilon if n > N 2 2n+1 2 Let No be the max of { 2 N , 2 N + 1 }. 1 2 Then if n > No, whether n is even or odd, | a - L | < epsilon. n 7. If Limit (a - a ) = 0, then Limit a exists and is finite. n->Infinity n n-1 n->Infinity n FALSE: Let S be the partial sums of the harmonic series. n /|\ Sn | Infinity. Let Limit S - S = Limit 1/n = 0 n->Infinity n n-1 n->Infinity 8. If {a } and {b } both diverge, then {a + b } diverges. n n n n FALSE: Let a be the harmonic series and let b be the negative harmonic series. n n both SUM a and SUM b diverge, but SUM a + b = 0 n n n n a n 9. If {a } converges, then {----} converges to 0. n n a n TRUE: If a -------> a, then ------ ---> 0 n n INF INF 2 10. If SUM a converges, so does SUM a n=1 n n=1 n n INF (-1) FALSE: SUM ------- converges by alternating series test. n=1 Sqrt[n] INF 1 SUM ----- diverges because it is the harmonic series. 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 TRUE: This is just the alternating series test. INF n 12. SUM (1/n) converges and has sum S satisfying 1 < S < 2. n=1 INF n INF 2 2 1 < SUM (1/n) <= SUM (1/n) = Pi /6 < 2 n=1 n=1 ------------------------------------------ Get["font.math"]; f[n_] := Sum[1.0/i^i,{i,1,n}] A = Table[f[n],{n,1,20}] p1 = ListPlot[A]; p2 = ParametricPlot[{0,y},{y,0,1}]; p3 = Plot[0,{x,0,20}]; p4 = Show[p1,p2,p3,PlotLabel->"P505 P12 SUM (1/n)^n"]; Display["p12.ps",p4]; ------------------------------------------------------ 13. If a series SUM a diverges, then its sequence of partial sums is unbounded. n FALSE: 1-1+1-1+1-1..... is a counterexample. 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 1 FALSE: ------ < 1 is a counterexample n 2 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 TRUE: These rational functions will always give r = 1 which is indecisive. 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 False: For a convergent p-series, the ratios always have limit 1. _ _ n INF | 1 | 17. SUM | 1 - --- | converges. n=1 |_ n _| FALSE: THE n-th term does not go to zero. It goes to 1/e. -------------------------------------- Get["font.math"]; f[n_] := Sum[ (1.0-1/i)^i,{i,1,n}] A = Table[f[n],{n,1,20}] p1 = ListPlot[A]; p2 = ParametricPlot[{0,y},{y,0,1}]; p3 = Plot[0,{x,0,20}]; p4 = Show[p1,p2,p3,PlotLabel->"P505 P17 SUM (1-1/n)^n"]; Display["p17.ps",p4]; -------------------------------------- INF 1 18. SUM ------------- converges. n=1 4 ln( n + 1) -------------------------------------- 1 FALSE: a = --------------- n 4 ln( n + 1 ) 1 b = ------- n ln(n) a 4 n ln(n) 1/n n +1 Limit --- = Limit ------ = Limit ----- = Limit ---- = 1/4 n->Inf b n->Inf 4 n->Inf 3 n->Inf 4 n ln(n +1) 4 n 4 n ------ 4 n + 1 INF INF SUM 1/ln(n) > SUM 1/n n=2 n=2 INF SUM 1/n is the harmonic series which diverges. n=1 INF SUM 1/ln(n) diverges by the comparison test. n=2 INF 4 SUM 1/ln(n + 1) diverges by the limit comparison test. n=1 ----------------------------------------------- Get["font.math"]; f[n_] := Sum[ 1.0/Log[i^4+1],{i,1,n}]; g[n_] := Sum[ 1.0/Log[n],{i,2,n}]; A = Table[f[n],{n,1,2000}]; B = Table[g[n],{n,2,2000}]; p1 = ListPlot[A]; p2 = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; p3 = ParametricPlot[{0,y},{y,0,1}]; p4 = Plot[0,{x,0,2000}]; p5 = Show[p1,p2,p3,p4, PlotLabel->"P505 P18 black:SUM 1.0/Log[n^4+1];red: SUM 1/Log[n]", PlotRange->All]; Display["p18.ps",p4]; -------------------------------------- INF n+1 19. SUM --------- converges. n=2 2 (n ln(n) ) INF n+1 INF 1 INF 1 TRUE: SUM ---------- = SUM ----------- + SUM ------------- n=2 2 n=2 2 n=2 2 2 (n ln(n) ) n (ln(n)) n (ln(n)) The second series on the right hand side is smaller than the p-series, p=2 so it converges. The first term on the right hand side is done by the integral test. _ _ INF INF 1 | - 1 | INT ---------- dx = | ------ | = 1/ln(2) x=2 2 |_ ln(x) _| x=2 x (ln(x)) Therefore the first term on the right hand side converges by the integral test. -------------------------------------- Get["font.math"]; f[n_] := Sum[1.0 (i+1)/(i Log[i])^2,{i,2,n}] A = Table[f[n],{n,2,200}] p1 = ListPlot[A]; p2 = ParametricPlot[{0,y},{y,0,1}]; p3 = Plot[0,{x,0,20}]; p4 = Plot[f[2]+1.78537,{x,0,200},PlotStyle->{RGBColor[1,0,0]}]; p5 = Plot[ +1.78537,{x,0,200},PlotStyle->{RGBColor[1,0,0]}]; p6 = Show[p1,p2,p3,p4,p5,PlotLabel->"P505 P19 SUM (i+1)/(i Log[i])^2", PlotRange->All]; Display["p19.ps",p6]; Integrate[(x+1)/(x Log[x])^2,{x,2,Infinity}] -------------------------------------- 2 n Pi Sin (------ ) INF 2 20. SUM ------------------ converges n=1 n FALSE: This is the odd terms of the harmonic series and it diverges. 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 TRUE: a converges by the comparison test and n+100 INF INF SUM a = S + SUM a which converges. n=1 n 99 n=1 n+100 22. If, for some c > 0, c a >= 1/n for all natural numbers n, n INF then SUM a diverges. n=1 n INF INF 1/c TRUE: SUM a > SUM ---- n=1 n n=1 n The right hand series is the harmonic series which diverges. The original series diverges by the comparison test. 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 FALSE: For an counterexample, let the original series be the alternating harmonic series. 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 True. This is just the negative form of the comparison test. 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 TRUE: This is the absolute convergence theorem. | INF n+1 1 99 n+1 1 | 27. | SUM (-1) --- - SUM (-1) --- | < 0.01 | n=1 n n=1 n | TRUE. The error is at most the first ommitted term which is 1/100. INF INF | | 28. If SUM a diverges then SUM | a | diverges. n=1 n n=1 | n | TRUE: This is the contrapositive of the absolute convergence theorem.