Page 1 11.4 Comparison Tests 777 Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known. 11.4 THEOREM 10 The Comparison Test Let be a series with no negative terms. (a) converges if there is a convergent series with for all for some integer N. (b) diverges if there is a divergent series of nonnegative terms with for all for some integer N. n 7 N, a n Ú d n gd n ga n n 7 N, a n â€¦ c n gc n ga n ga n Proof In Part (a), the partial sums of are bounded above by They therefore form a nondecreasing sequence with a limit In Part (b), the partial sums of are not bounded from above. If they were, the par- tial sums for would be bounded by and would have to converge instead of diverge. EXAMPLE 1 Applying the Comparison Test (a) The series diverges because its nth term is greater than the nth term of the divergent harmonic series. 5 5n - 1 = 1 n - 1 5 7 1 n a q n = 1 5 5n - 1 gd n M * = d 1 + d 2 + Á + d N + a q n = N + 1 a n gd n ga n L â€¦ M. M = a 1 + a 2 + Á + a N + a q n = N + 1 c n . ga n HISTORICAL BIOGRAPHY Albert of Saxony (ca. 1316â€“1390) 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 777 Page 2 11.4 Comparison Tests 777 Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known. 11.4 THEOREM 10 The Comparison Test Let be a series with no negative terms. (a) converges if there is a convergent series with for all for some integer N. (b) diverges if there is a divergent series of nonnegative terms with for all for some integer N. n 7 N, a n Ú d n gd n ga n n 7 N, a n â€¦ c n gc n ga n ga n Proof In Part (a), the partial sums of are bounded above by They therefore form a nondecreasing sequence with a limit In Part (b), the partial sums of are not bounded from above. If they were, the par- tial sums for would be bounded by and would have to converge instead of diverge. EXAMPLE 1 Applying the Comparison Test (a) The series diverges because its nth term is greater than the nth term of the divergent harmonic series. 5 5n - 1 = 1 n - 1 5 7 1 n a q n = 1 5 5n - 1 gd n M * = d 1 + d 2 + Á + d N + a q n = N + 1 a n gd n ga n L â€¦ M. M = a 1 + a 2 + Á + a N + a q n = N + 1 c n . ga n HISTORICAL BIOGRAPHY Albert of Saxony (ca. 1316â€“1390) 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 777 778 Chapter 11: Infinite Sequences and Series (b) The series converges because its terms are all positive and less than or equal to the correspon- ding terms of The geometric series on the left converges and we have The fact that 3 is an upper bound for the partial sums of does not mean that the series converges to 3. As we will see in Section 11.9, the series con- verges to e. (c) The series converges. To see this, we ignore the first three terms and compare the remaining terms with those of the convergent geometric series The term of the truncated sequence is less than the corresponding term of the geometric se- ries. We see that term by term we have the comparison, So the truncated series and the original series converge by an application of the Com- parison Test. The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which is a rational function of n. a n 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á â€¦ 1 + 1 2 + 1 4 + 1 8 + Á 1>2 n 1>s2 n +2nd g q n=0 s1>2 n d. 5 + 2 3 + 1 7 + 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á + 1 2 n +2n + Á g q n=0 s1>n!d 1 + a q n = 0 1 2 n = 1 + 1 1 - s1>2d = 3. 1 + a q n = 0 1 2 n = 1 + 1 + 1 2 + 1 2 2 + Á . a q n = 0 1 n! = 1 + 1 1! + 1 2! + 1 3! + Á THEOREM 11 Limit Comparison Test Suppose that and for all (N an integer). 1. If then and both converge or both diverge. 2.If and converges, then converges. 3.If and diverges, then diverges. ga n gb n lim n: q a n b n = q ga n gb n lim n: q a n b n = 0 gb n ga n lim n: q a n b n = c 7 0, n Ú N b n 7 0 a n 7 0 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 778 Page 3 11.4 Comparison Tests 777 Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known. 11.4 THEOREM 10 The Comparison Test Let be a series with no negative terms. (a) converges if there is a convergent series with for all for some integer N. (b) diverges if there is a divergent series of nonnegative terms with for all for some integer N. n 7 N, a n Ú d n gd n ga n n 7 N, a n â€¦ c n gc n ga n ga n Proof In Part (a), the partial sums of are bounded above by They therefore form a nondecreasing sequence with a limit In Part (b), the partial sums of are not bounded from above. If they were, the par- tial sums for would be bounded by and would have to converge instead of diverge. EXAMPLE 1 Applying the Comparison Test (a) The series diverges because its nth term is greater than the nth term of the divergent harmonic series. 5 5n - 1 = 1 n - 1 5 7 1 n a q n = 1 5 5n - 1 gd n M * = d 1 + d 2 + Á + d N + a q n = N + 1 a n gd n ga n L â€¦ M. M = a 1 + a 2 + Á + a N + a q n = N + 1 c n . ga n HISTORICAL BIOGRAPHY Albert of Saxony (ca. 1316â€“1390) 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 777 778 Chapter 11: Infinite Sequences and Series (b) The series converges because its terms are all positive and less than or equal to the correspon- ding terms of The geometric series on the left converges and we have The fact that 3 is an upper bound for the partial sums of does not mean that the series converges to 3. As we will see in Section 11.9, the series con- verges to e. (c) The series converges. To see this, we ignore the first three terms and compare the remaining terms with those of the convergent geometric series The term of the truncated sequence is less than the corresponding term of the geometric se- ries. We see that term by term we have the comparison, So the truncated series and the original series converge by an application of the Com- parison Test. The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which is a rational function of n. a n 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á â€¦ 1 + 1 2 + 1 4 + 1 8 + Á 1>2 n 1>s2 n +2nd g q n=0 s1>2 n d. 5 + 2 3 + 1 7 + 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á + 1 2 n +2n + Á g q n=0 s1>n!d 1 + a q n = 0 1 2 n = 1 + 1 1 - s1>2d = 3. 1 + a q n = 0 1 2 n = 1 + 1 + 1 2 + 1 2 2 + Á . a q n = 0 1 n! = 1 + 1 1! + 1 2! + 1 3! + Á THEOREM 11 Limit Comparison Test Suppose that and for all (N an integer). 1. If then and both converge or both diverge. 2.If and converges, then converges. 3.If and diverges, then diverges. ga n gb n lim n: q a n b n = q ga n gb n lim n: q a n b n = 0 gb n ga n lim n: q a n b n = c 7 0, n Ú N b n 7 0 a n 7 0 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 778 11.4 Comparison Tests 779 Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b). Since there exists an integer N such that for all n Thus, for If converges, then converges and converges by the Direct Compari- son Test. If diverges, then diverges and diverges by the Direct Com- parison Test. EXAMPLE 2 Using the Limit Comparison Test Which of the following series converge, and which diverge? (a) (b) (c) Solution (a)Let For large n, we expect to behave like since the leading terms dominate for large n, so we let Since and diverges by Part 1 of the Limit Comparison Test. We could just as well have taken but 1 n is simpler. > b n = 2>n, ga n lim n: q a n b n = lim n: q 2n 2 + n n 2 + 2n + 1 = 2, a q n = 1 b n = a q n = 1 1 n diverges b n = 1>n. 2n>n 2 = 2>n a n a n = s2n + 1d>sn 2 + 2n + 1d. 1 + 2 ln 2 9 + 1 + 3 ln 3 14 + 1 + 4 ln 4 21 + Á = a q n = 2 1 + n ln n n 2 + 5 1 1 + 1 3 + 1 7 + 1 15 + Á = a q n = 1 1 2 n - 1 3 4 + 5 9 + 7 16 + 9 25 + Á = a q n = 1 2n + 1 sn + 1d 2 = a q n = 1 2n + 1 n 2 + 2n + 1 ga n gsc>2db n gb n ga n gs3c>2db n gb n a c 2 bb n 6 a n 6 a 3c 2 bb n . c 2 6 a n b n 6 3c 2 , - c 2 6 a n b n - c 6 c 2 , n 7 N, n 7 N Q ` a n b n - c` 6 c 2 . c>2 7 0, Limit definition with and replaced by a n >b n a n P= c>2, L = c, 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 779 Page 4 11.4 Comparison Tests 777 Comparison Tests We have seen how to determine the convergence of geometric series, p-series, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known. 11.4 THEOREM 10 The Comparison Test Let be a series with no negative terms. (a) converges if there is a convergent series with for all for some integer N. (b) diverges if there is a divergent series of nonnegative terms with for all for some integer N. n 7 N, a n Ú d n gd n ga n n 7 N, a n â€¦ c n gc n ga n ga n Proof In Part (a), the partial sums of are bounded above by They therefore form a nondecreasing sequence with a limit In Part (b), the partial sums of are not bounded from above. If they were, the par- tial sums for would be bounded by and would have to converge instead of diverge. EXAMPLE 1 Applying the Comparison Test (a) The series diverges because its nth term is greater than the nth term of the divergent harmonic series. 5 5n - 1 = 1 n - 1 5 7 1 n a q n = 1 5 5n - 1 gd n M * = d 1 + d 2 + Á + d N + a q n = N + 1 a n gd n ga n L â€¦ M. M = a 1 + a 2 + Á + a N + a q n = N + 1 c n . ga n HISTORICAL BIOGRAPHY Albert of Saxony (ca. 1316â€“1390) 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 777 778 Chapter 11: Infinite Sequences and Series (b) The series converges because its terms are all positive and less than or equal to the correspon- ding terms of The geometric series on the left converges and we have The fact that 3 is an upper bound for the partial sums of does not mean that the series converges to 3. As we will see in Section 11.9, the series con- verges to e. (c) The series converges. To see this, we ignore the first three terms and compare the remaining terms with those of the convergent geometric series The term of the truncated sequence is less than the corresponding term of the geometric se- ries. We see that term by term we have the comparison, So the truncated series and the original series converge by an application of the Com- parison Test. The Limit Comparison Test We now introduce a comparison test that is particularly useful for series in which is a rational function of n. a n 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á â€¦ 1 + 1 2 + 1 4 + 1 8 + Á 1>2 n 1>s2 n +2nd g q n=0 s1>2 n d. 5 + 2 3 + 1 7 + 1 + 1 2 +21 + 1 4 +22 + 1 8 +23 + Á + 1 2 n +2n + Á g q n=0 s1>n!d 1 + a q n = 0 1 2 n = 1 + 1 1 - s1>2d = 3. 1 + a q n = 0 1 2 n = 1 + 1 + 1 2 + 1 2 2 + Á . a q n = 0 1 n! = 1 + 1 1! + 1 2! + 1 3! + Á THEOREM 11 Limit Comparison Test Suppose that and for all (N an integer). 1. If then and both converge or both diverge. 2.If and converges, then converges. 3.If and diverges, then diverges. ga n gb n lim n: q a n b n = q ga n gb n lim n: q a n b n = 0 gb n ga n lim n: q a n b n = c 7 0, n Ú N b n 7 0 a n 7 0 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 778 11.4 Comparison Tests 779 Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b). Since there exists an integer N such that for all n Thus, for If converges, then converges and converges by the Direct Compari- son Test. If diverges, then diverges and diverges by the Direct Com- parison Test. EXAMPLE 2 Using the Limit Comparison Test Which of the following series converge, and which diverge? (a) (b) (c) Solution (a)Let For large n, we expect to behave like since the leading terms dominate for large n, so we let Since and diverges by Part 1 of the Limit Comparison Test. We could just as well have taken but 1 n is simpler. > b n = 2>n, ga n lim n: q a n b n = lim n: q 2n 2 + n n 2 + 2n + 1 = 2, a q n = 1 b n = a q n = 1 1 n diverges b n = 1>n. 2n>n 2 = 2>n a n a n = s2n + 1d>sn 2 + 2n + 1d. 1 + 2 ln 2 9 + 1 + 3 ln 3 14 + 1 + 4 ln 4 21 + Á = a q n = 2 1 + n ln n n 2 + 5 1 1 + 1 3 + 1 7 + 1 15 + Á = a q n = 1 1 2 n - 1 3 4 + 5 9 + 7 16 + 9 25 + Á = a q n = 1 2n + 1 sn + 1d 2 = a q n = 1 2n + 1 n 2 + 2n + 1 ga n gsc>2db n gb n ga n gs3c>2db n gb n a c 2 bb n 6 a n 6 a 3c 2 bb n . c 2 6 a n b n 6 3c 2 , - c 2 6 a n b n - c 6 c 2 , n 7 N, n 7 N Q ` a n b n - c` 6 c 2 . c>2 7 0, Limit definition with and replaced by a n >b n a n P= c>2, L = c, 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 779 (b)Let For large n,we expect to behave like so we let Since and converges by Part 1 of the Limit Comparison Test. (c)Let For large n, we expect to behave like which is greater than 1 nfor so we take Since and diverges by Part 3 of the Limit Comparison Test. EXAMPLE 3Does converge? Solution Because ln n grows more slowly than for any positive constant c (Section 11.1, Exercise 91), we would expect to have for n sufficiently large. Indeed, taking and we have Since (a p-series with ) converges, converges by Part 2 of the Limit Comparison Test. ga n p 7 1 gb n = gs1>n 5>4 d = lim n: q 4 n 1>4 = 0. = lim n: q 1>n s1>4dn -3>4 lim n: q a n b n = lim n: q ln n n 1>4 b n = 1>n 5>4 , a n = sln nd>n 3>2 ln n n 3>2 6 n 1>4 n 3>2 = 1 n 5>4 n c a q n = 1 ln n n 3>2 ga n = q , lim n: q a n b n = lim n: q n + n 2 ln n n 2 + 5 a q n = 2 b n = a q n = 2 1 n diverges b n = 1>n. n Ú 3, > sn ln nd>n 2 = sln nd>n, a n a n = s1 + n ln nd>sn 2 + 5d. ga n = 1, = lim n: q 1 1 - s1>2 n d lim n: q a n b n = lim n: q 2 n 2 n - 1 a q n = 1 b n = a q n = 1 1 2 n converges b n = 1>2 n . 1>2 n , a n a n = 1>s2 n - 1d. 780 Chapter 11: Infinite Sequences and Series lâ€™Hôpitalâ€™ s Rule 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 780Read More

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!