Week 9 lecture - The Integral and Comparison Tests for...

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Unformatted text preview: The Integral and Comparison Tests for Series Integral Test Motivation Seems like the series is convergent! We can justify that this series is convergent geometrically looking at the area under the function y=1/x2 Integral test motivation Improper integral What do we know about the integral Z 1 1 1 dx = lim t!1 x2 Z an = t 1 1 dx = lim t!1 x2 1 for n n ? 1 x t =1 1 1 orem to calculate the surface integral Z ~ ~ Series converges!!! F · dS, that is calculate the ﬂux of Another example: Z 1 1 1 p dx = lim t!1 x Z t 1 1 p dx = lim 2x1/2 t!1 x t 1 =1 1 for n 1 n Z Series diverges!!! ~ ~ ce Theorem to calculate the surface integral F · dS, that is calculate the ﬂux of an = S REMARK: Don’t use the Integral Test to evaluate series, because in general The Integral Test ∞ n=1 ∞ an ̸= f (x)dx 1 EXAMPLES: 1. ∞ n=1 divergent by the p-test for improper integrals, since p = 1 ≤ 1. 2. ∞ n=1 1 n1/2 is divergent, because f (x) = 1 x1/2 3. n=1 1 1 is convergent, because f (x) = 2 is continuous, positive, decreasing and n2 x convergent by the p-test for improper integrals, since p = 2 > 1. 1 dx is x 1 ∞ is continuous, positive, decreasing and is divergent by the p-test for improper integrals, since p = 1/2 ≤ 1. ∞ ∞ 1 1 is divergent, because f (x) = is continuous, positive, decreasing and n x 1 x1/2 dx 1 ∞ 1 dx is x2 1 REMARK 1: When we use the Integral Test it is not necessary to start the series or the integral at n = 1. For instance, in testing the series ∞ n=5 1 n+1 ∞ we use 1 dx x+1 5 REMARK 2: It is not necessary that f be always decreasing. It has to be ultimately decreasing, that is, decreasing for x larger than some number N. EXAMPLE: Determine whether the series ∞ ln n converges or diverges. n Practice: Section 8.3 The Integral and Comparison Tests EXAMPLE: Determine whether the series 2010 ∞ n=1 ln n converges or diverges. n ln x Solution: The function f (x) = is positive and continuous for x > 1. Howev x the graph of this function we conclude that f is not decreasing. At the same time one can show that this function is ultimately decreasing. In fa ′ f (x) = ln x ′ = (ln x)′ · x − ln x · x′ = 1 x · x − ln x · 1 = 1 − ln x Practice: Section 8.3 The Integral and Comparison Tests 2010 Section 8.3 The Integral and Comparison Tests ∞ 2010 Kiryl Tsishchanka ln n converges or diverges. EXAMPLE: Determine whether the series converges or diverges. n n n=1 n=1 ln x ln x Solution: The function f (x) = is positive and continuous for x > 1. However, looking at Solution: of this function we conclude that f is not decreasing. The function x (x) = f is positive and continuous for x > 1. Howev the graph x the graph of this function we conclude that f is not decreasing. ∞ EXAMPLE: Determine whether ln n series the At the same time one can show that this function is ultimately decreasing. In fact, ′ f (x) = ln x x ′ (ln x)′ · x − ln x · x′ = = x2 1 x · x − ln x · 1 1 − ln x = 2 x x2 Note that 1 − ln x < 0 for all suﬃciently large x which means that f ′ (x) < 0 and therefore f is ultimately decreasing. So, we can apply the Integral Test: ∞ t t At the same time one canlnshow thatx)this function is ultimately decreasing. In fa ln x x (ln 2 (ln t)2 x 1 dx = lim x t→∞ 1 dx = lim t→∞ ′ 2 = lim 1 t→∞ 2 =∞ ln x ∞ (ln x)′ · x − ln x · x′ ′ f (x) = = ln= n Since this integral diverges, the series x n=1 n also diverges. x2 1 x · x − ln x · 1 1 − ln x = x2 x2 Practice: Since this integral diverges, the series ∞ n=1 ln n also diverges. n EXAMPLE: Determine whether the series ∞ n=2 1 converges or diverges. n ln n 1 Solution: The function f (x) = is continuous, positive and decreasing on [2, ∞), t x ln x we can apply the Integral Test: ∞ 2 t 1 dx = lim t→∞ x ln x 1 dx = lim ln(ln x)]t = lim [ln(ln t) − ln(ln 2)] = ∞ 2 t→∞ t→∞ x ln x 2 Since this integral diverges, the series ∞ n=2 1 also diverges. n ln n EXAMPLE: Determine whether the series ∞ n=2 1 converges or diverges. n ln2 n ln x (ln x)2 dx = lim ∞ t→∞ x 2 ln n ln x dx = lim t→∞ x Practice: 1 1 Since this integral diverges, the series Since this integral diverges, the series ∞ n=1 n n=1 1 (ln t)2 = lim =∞ t→∞ 2 also diverges. ln n also diverges. n ∞ 1 converges or diverges. n ln n EXAMPLE: Determine whether the series ∞ 1 n=2 converges or diverges. n ln n EXAMPLE: Determine whether the series 1 n=2 Solution: The function f (x) = 1 x ln x is continuous, positive and decreasing on [2, ∞), t Solution: The we can applyfunction f (x) = Test: is continuous, positive and decreasing on [2, ∞), therefore the Integral x ln x we can apply the Integral Test: ∞ ∞ 22 t t 1 t 1 1 dx = lim 1 dx ln(ln x)]2 = lim [ln(ln − ln(ln 2)] = ∞ lim dx = limt→∞ dx = lim = t→∞ tln(ln x)]2 =t)lim [ln(ln t) − ln(ln 2)] = ∞ t→∞ t→∞ t→∞ xx ln x ln x x ln xx ln xt→∞ 2 2 ∞ 1∞ 1 Since this integral diverges, the series ln n also diverges. diverges. also n n=2 Since this integral diverges, the series n=2 EXAMPLE: Determine whether the series ∞ n ln n 1 ∞ converges or diverges. 1 n ln2 n n=2 EXAMPLE: Determine whether the series n=2 2 n ln2 n converges or diverges. Practice: Section 8.3 The Integral and Comparison Tests 201 EXAMPLE: Determine whether the series ∞ n=2 1 converges or diverges. 2 n ln n 1 Solution: The function f (x) = is continuous, positive and decreasing on [ 2 x ln x we can apply the Integral Test: ∞ 2 1 dx = lim t→∞ x ln2 x t 2 1 1 dx = lim − t→∞ ln x x ln2 x Since this integral converges, the series mentioned before, ∞ n=2 ∞ n=2 t 2 1 1 = lim − + t→∞ ln t ln 2 1 also converges. We also note 2 n ln n 1 1 ̸= . 2 ln 2 n ln n THEOREM (p-Test): The p-series ∞ n=1 1 is convergent if p > 1 and divergent p n Practice: Section 8.3 The Integral and Comparison Tests 201 Section 8.3 The Integral and Comparison Tests EXAMPLE: Determine whether the series ∞ ∞ 2010 Kiryl Tsishchanka 1 converges or diverges. 2 n ln n 1 n=2 converges or diverges. 2 n ln n EXAMPLE: Determine whether the series 1 n=2 Solution: The function f (x) = is continuous, positive and decreasing on [ 2 1 x ln x Solution: The function f (x) = is continuous, positive and decreasing on [2, ∞), therefore 2 we can apply the Integral ln x x Test: we can apply the Integral Test: ∞ 2 ∞ t t t t 1 1 1 dx = lim1 1 lim − 1 1 = lim − 1 1 + 1 1 dx = lim dx = 2limdx = − = lim − + = 2 2 t→∞ t→∞ ln ln ln 2 ln 2 t x x x t→∞ t→∞ 2 x x ln t→∞ ln x 2 t→∞ x ln2t ln 2 ln x x ln x ln 2 2 2 ∞ 1 ∞ 1 Since this integral converges, the series also converges. We also note that, as it was 2 Since this integral converges, the series n also converges. We also note n ln 2 n=2 mentioned before, ∞ mentioned before, n=2 n=2 1 1 ∞ ̸=1 . ln 2 n ln2 n n=2 n ln n 1 ̸= . 2 ln 2 n ln n THEOREM (p-Test): The p-series ∞ n=1 1 is convergent if p > 1 and divergent if p ≤ 1. np ∞ THEOREM (p-Test): cases: Proof: We distinguish threeThe p-series 1 1 is convergent if p > 1 and divergent p ∞n n=1 1 Since this integral converges, the series p-test mentioned before, ∞ n=2 n=2 n ln2 n also converges. We also note that, as it w 1 1 ̸= . 2 ln 2 n ln n THEOREM (p-Test): The p-series ∞ n=1 1 is convergent if p > 1 and divergent if p ≤ 1. np Proof: We distinguish three cases: 1 Case I: If p < 0, then lim p = ∞, therefore n→∞ n ∞ n=1 1 diverges by the Divergence Test. np ∞ 1 1 1 1 Case II: If p = 0, then lim p = lim 0 = lim = 1, therefore diverges by t n→∞ n n→∞ n n→∞ 1 np n=1 Divergence Test. 1 Case III: If p > 0, then the function f (x) = p is continuous, positive and decreasing x ∞ 1 [1, ∞), therefore we can apply the Integral Test by which is convergent if and only np n=1 ∞ the improper integral 1 1 dx is convergent. But xp if p ≤ 1 by the p-test for improper integrals. ∞ 1 dx is convergent if p > 1 and diverge xp 1 REMARK: As before, when we use the p-Test it is not necessary to start the series at n = t→∞ Since this integral converges, x ln series the x x ln x 2 2 ∞ t→∞ converges. We alsoln 2 that, as it w note ln x alsot→∞ ln t ln 2 2 n ln2 n n=2 1 ∞ ∞ Since this integral converges, the series also converges. We also note that, as it was 1 1 n ln21 also converges. We also note that, as it was n Since this integral n=2 mentioned before, converges, the series. ̸= 2 2 ∞ ln 2 n=2 n ln n n 1 ln n 1 ∞ mentioned before, n=2 2 1 ̸= . ln n ln n ̸= 21 . mentioned before, n=2 ln 2 ∞ 1 n ln2 n n=2 p-test THEOREM (p-Test): The p-series ∞ is convergent if p > 1 and divergent if p ≤ 1. p 1 ∞n=1 n THEOREM (p-Test): The p-series is convergent if p > 1 and divergent if p ≤ 1. p1 THEOREM (p-Test): The p-series n p is convergent if p > 1 and divergent if p ≤ 1. n=1 Proof: We distinguish three cases: n n=1 ∞ Proof: We distinguish three cases: 1 1 ∞ Proof: We < 0, then lim Case I: If p distinguish three cases: ∞, therefore1 diverges by the Divergence Test. 1 p = n→∞ = n np Case I: If p < 0, then lim p 1 ∞, therefore ∞ p 1diverges by the Divergence Test. n→∞ n Case I: If p < 0, then lim p = ∞, therefore n n=1diverges by the Divergence Test. ∞ n=1 p n→∞ n ∞ 1 1n=1 n1 1 1 1 Case II: IfIfp p = 0, then lim =p lim lim= 0lim lim 1, therefore ∞ 1 diverges by the by t II: = 1 = = = 1, therefore diverges Case = 0, then lim p 1n 1 1 n→∞ n→∞ n n→∞ 1 np by the 01 p n→∞ n n→∞ n n→∞ 1 Case II: If p = 0, then lim p = lim 0 = lim = 1, thereforen=1 n p n=1 diverges n→∞ n n→∞ n n→∞ 1 n Divergence Test. n=1 Divergence Test. Divergence Test. 1 1 Case III: p > 0, then the function f (x) (x) 1 is p is continuous, and decreasing on Case III: IfIf p > 0, then the function f= p = continuous, positive positive and decreasing x Case III: If p > 0, then the function f (x) =x p is continuous, positive and decreasing on ∞ ∞ x ∞1 1 [1, ∞), therefore wewe can apply the Integral by which which1 is convergent if and only if [1,[1, ∞), therefore we can apply the Integral Test by which np is convergent if and onlyand only ∞), therefore can apply the Integral Test Test by is convergent if if np p n=1 n n=1 ∞ ∞ ∞ ∞ n=1 ∞ ∞ 1 1 1 1convergent if p > 1 1 1 the improper integral dx is convergent. But the improper integral xp p p dx is convergent. But dx isis dx is convergent1and divergent diverge the improper integral dx is convergent. But xp p dx p convergent if p > if p > 1 and and divergent xx x x 1 1 1 1 if p ≤ 1 by the p-test for 11 improper integrals. if ≤≤ 1 bythe p-test for improper integrals. p 1 by the p-test for improper integrals. if p REMARK: AsAs before, when we use the p-Test it is not necessary to start theseries at nn= 1. REMARK: before, when we use the p-Test it is not necessary to start the series at = 1. REMARK: As before, when we use the p-Test it is not necessary to start the series at n = EXAMPLE: Determine whether the following series converge or diverge: Practice: Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchan EXAMPLE: Determine whether the series ∞ n=1 1 converges or diverges. n+1 REMARK: Note that we CAN’T apply the p-test directly! 1 is continuous, positive and decreasing on [1, ∞), ther x+1 fore we can apply the Integral Test: Solution 1: The function f (x) = ∞ t 1 dx = lim t→∞ x+1 1 1 dx = lim ln(x + 1)]t = lim [ln(t + 1) − ln 2] = ∞ 1 t→∞ t→∞ x+1 1 Since this integral diverges, the series Solution 2: We have ∞ n=1 ∞ n=1 ∞ 1 also diverges. n+1 1 = n+1 ∞ n=2 1 n ∞ 1 1 Since diverges by the p-test with p = 1, it follows that also diverges, sin n n+1 n=2 n=1 convergence or divergence is unaﬀected by deleting a ﬁnite number of terms. Practice: Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchan Section 8.3 The Integral and Comparison Tests EXAMPLE: Determine whether the series ∞ EXAMPLE: Determine whether the series n=1 ∞ 2010 Kiryl Tsishchanka 1 converges or diverges. n+1 1n=1 converges or diverges. n+1 REMARK: Note that we CAN’T apply the p-test directly! REMARK: Note that we CAN’T apply the p-test directly! 1 Solution 1: The function f (x)(x) = is continuous, positive and decreasing on [1, ∞), there- [1, ∞), ther function f = 1 is continuous, positive and decreasing on Solution 1: The x+1 x+1 fore we can apply the Integral Test: fore we can apply the Integral Test: ∞ ∞ 1 t t 1 1 1 dx = lim 1 dx = lim ln(x + 1)]t = lim [ln(t + 1) − ln 2] = ∞ 1 t→∞ t→∞ t→∞ x + 1 dx = lim x + 1 dx = lim ln(x + 1)]t = lim [ln(t + 1) − ln 2] 1 x+1 1 x+1 t→∞ 1 1 Since this integral diverges, the series ∞ t→∞ Solution 2: We have ∞ n=1 n+1 ∞ n=1 1 = n+1 ∞ n=2 1 n also diverges. ∞ 1 1 ∞ = 1 diverges by the p-test with p = 1, n + 1 it follows that n n=1 n=2 1 also diverges, since n n+1 n=2 n=1 ∞ convergence or divergence is unaﬀected by deleting a ﬁnite number of terms. ∞ 1 1 Since Since =∞ 1 ∞ also diverges. n+1 1 n=1 Since this 2: We have Solution integral diverges, the series ∞ t→∞ diverges by the p-test with p = 1, it follows that also diverges, sin n n+1 The Comparison Tests n=2 n=1 convergence or divergence is unaﬀected by deleting a ﬁnite number of terms. The comparison tests!!! Slogan: Smaller than convergent is convergent!!! Bigger than divergent is divergent!!! Motivation: n=1 n=2 ∞ ∞ 1 1 Since diverges by the p-test with p = 1, it follows that also diverges, since n n+1 n=2 n=1 convergence or divergence is unaﬀected by deleting a ﬁnite number of terms. The Comparison Test The Comparison Tests THE COMPARISON TEST: Suppose that an and (a) If bn is convergent and an ≤ bn for all n, then (b) If bn is divergent and an ≥ bn for all n, then bn are series with positive terms. an is also convergent an is also divergent EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge. ∞ ∞ ln n 1 √ (a) (b) 3 n n−1 n=1 n=2 ln n 1 Solution: Since > for n > e and n n ∞ n=1 ∞ n=1 1 diverges by the p-test with p = 1, it follows that n ln n 1 1 also diverges. Similarly, since √ > √ and 3 3 n n−1 n p = 1/3, it follows that ∞ n=2 √ 3 ∞ n=2 1 √ diverges by the p-test with 3 n 1 also diverges. n−1 EXAMPLE: Use the Comparison Test to determine whether the following series converge or (a) If bn is convergent and an ≤ bn for all n, then Practice: and a b is divergent (b) If n n ≥ bn for all n, then an is also convergent an is also divergent EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge. ∞ ∞ ln n 1 √ (a) (b) 3 n n−1 n=1 n=2 ln n 1 Solution: Since > for n > e and n n ∞ n=1 ∞ n=1 1 diverges by the p-test with p = 1, it follows that n ln n 1 1 also diverges. Similarly, since √ > √ and 3 3 n n−1 n p = 1/3, it follows that ∞ n=2 √ 3 ∞ n=2 1 √ diverges by the p-test with 3 n 1 also diverges. n−1 EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge: ∞ ∞ 1 1 (a) (b) n2 + n + 1 n2 − 1 n=1 n=2 4 THE COMPARISON TEST: Suppose that (a) If (b) If (b) If bn are series with positive terms. bn is convergent and an ≤ bn for all n, then Practice: and a b is divergent (a) If an and bn is convergent and an ≤ bn for all n, then ≥ b for all n, then n n n bn is divergent and an ≥ bn for all n, then an is also convergent an is also convergent a is also divergent an is n also divergent EXAMPLE: Use the Comparison Test to to determine whether following series series converge or EXAMPLE: Use the Comparison Test determine whether the the following converge or diverge. diverge. ∞ ∞ ∞ ∞ 1 lnln n n (a) (b) √ 1 √ (a) (b) 3 3 n−1 nn n n=1 n=2− 1 n=1 Solution: Since Solution: Since ∞ ∞ n=2 ∞ ∞ ln ln nn 1 1 1 1 > for > e e and diverges by the p-test p = p = 1, it follows > for n n > and diverges by the p-test with with1, it follows that that nn nn n n n=1 n=1 ∞ ∞ ln n 1 1 1 1 1 1 also diverges. Similarly, since √ √ > √ √ and √ diverges by the p-test p-test with and √ diverges by the with also diverges. Similarly, since n3− 1 >n 3 3 3 3 n n 3n n−1 n n=2 n=2 n=1 n n=1 p = 1/3, it follows that p = 1/3, it follows that ∞∞ n=2 1 √√ 1 also diverges. 3 n − − 1 also diverges. 3 n1 n=2 EXAMPLE: Use the Comparison Test to determine whether the following series converge or EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge: ∞ ∞ diverge: 1 1 ∞ ∞ (a) (b) 1 1 (a)n=1 n2 + n + 1 (b)n2 − 1 2 n=2 n2 + n + 1 n −1 n=1 n=2 4 4 Practice: Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge: (a) ∞ n=1 1 n2 + n + 1 Solution: 1 1 (a) Since 2 < 2 and n +n+1 n ∞ n=1 (b) ∞ n=2 ∞ n=1 1 n2 − 1 1 converges by the p-test with p = 2, it follows that n2 1 also converges. n2 + n + 1 1 2 (b) Since 2 < 2 for n ≥ 2 and n −1 n follows that ∞ n=2 ∞ n=2 ∞ 2 1 =2 converges by the p-test with p = 2, it n2 n2 n=2 1 also converges. n2 − 1 THE LIMIT COMPARISON TEST: Suppose that terms. If an lim =c an and bn are series with positive Practice: Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka Section 8.3 The Integral and Comparison Tests 2010 Kiryl Tsishchanka EXAMPLE: Use the Comparison Test to determine whether the following series converge or diverge: EXAMPLE: Use the Comparison Test to determine whether the following series converge or ∞ ∞ diverge: 1 1 (a) ∞ 2 1 (b)∞ 12 (a) n=1 n2 + n + 1 (b) n=22n − 1 n +n+1 n −1 n=1 n=2 Solution: ∞ Solution: 1 1 1 (a) Since 2 1 < 12 and ∞ 1 2 converges by the p-test with p = 2, it follows that (a) Since n 2 + n + 1 < n2 and n=1 n converges by the p-test with p = 2, it follows that n +n+1 n n2 ∞ n=1 1 ∞ also converges. 1 n22+ n + 1 also converges. n=1 n + n + 1 n=1 ∞ ∞ 1 2 2 1 ∞ ∞ 2 =2 1 converges by the p-test with p = 2, it (b) Since 2 1 < 22 for n ≥ 2 and (b) Since n 2 − 1 < n2 for n ≥ 2 and n2 = 2 n=2 2 2converges by the p-test with p = 2, it n n=2n2 n −1 n n n=2 n=2 ∞ ∞ 1 1 follows that also converges. follows that n2 − 1 also converges. 2−1 n=2 n n=2 THE LIMIT COMPARISON TEST: Suppose that a and THE LIMIT COMPARISON TEST: Suppose that an n and terms. If terms. If aan lim n = c n are series with positive bnbare series with positive n=1 ∞ ∞ 1 2 2 1 (b) Since 2 < 2 for n ≥ 2 and ∞ 2 = 2 ∞ 2 converges by the p-test with p = 2, it n −1 n2 n2 n1 1 n=2 n=2 (b) Since 2 ∞ < 2 for n ≥ 2 and =2 converges by the p-test with p = 2, it n −1 1n n2 n2 n=2 follows that ∞ 2 also converges. n=2 n −1 1 n=2 follows that also converges. n2 − 1 n=2 THE LIMIT COMPARISON TEST: Suppose that an and bn are series with positive terms. If THE LIMIT COMPARISON TEST: Supposen that an and bn are series with positive a lim =c terms. If n→∞ bn an lim both series converge or both diverge. where c is a ﬁnite number and c > 0, then n→∞ bn = c either The Limit Comparison Test ∞ where c is a ﬁnite number and c > 0, then either both series converge or both diverge. 1 EXAMPLE: Use the Limit Comparison Test to determine whether converges or n2 ∞ −1 n=2 1 diverges. EXAMPLE: Use the Limit Comparison Test to determine whether converges or n2 − 1 n=2 1 1 diverges. Solution: Put an = 2 , bn = 2 . Then n −1 n 1 1 Solution: Put an = 2 , bn = 12 . Then n −1 n 2−1 an n2 1 1 n c = lim = lim = lim 2 = lim = =1 11 n→∞ bn n→∞ n→∞ n − 1 n→∞ 1 − 1 1−0 2 2 n n 2 an 1 1 n2 − 1 = lim n c = lim = lim = lim = = ∞ ∞ 1 1 n→∞ 1 n n→∞ n→∞ n2 − 1 n→∞ 1 − 1 1 b 1−0 n2 n2 Since c = 1 and converges by the p-test with p = 2, it follows that also n2 n2 − 1 n=2 n=2 ∞ ∞ converges. 1 1 Since c = 1 and converges by the p-test with p = 2, it follows that also n2 n2 − 1 n=2 n=2 EXAMPLE: Use the Limit Comparison Test to determine whether the following series converge converges. ∞ n=1 2 1 1 (b) Since 2 also converges. ≥ 2 and < 2 for n n2 n2 + nn 1 1 +− 1 n=1 ∞ ∞ 2 1 = 2 ∞ 2 converges by the p-test with p = 2, it ∞ n2 n1 2 n=2 n=2 (b) Since 2 ∞ < 2 for n ≥ 2 and =2 converges by the p-test with p = 2, it n −1 1n n2 ∞ n=2 n2 ∞ follows that 1 ∞ 2 2 also converges. n=2 2 1 (b) Since 2 < 1 for =2 converges by the p-test with p = 2, it n − 1 n ≥ 2 and n=2 n −1 n2 n2 n2 n=2 n=2 follows that ∞ also converges. 2 n1 − 1 follows that n=2 2 also converges. THE LIMIT COMPARISON TEST: Suppose that an and bn are series with positive n −1 terms. If n=2 THE LIMIT COMPARISON TEST: Supposen that an and bn are series with positive a lim THE LIMIT COMPARISON TEST: Suppose that = c an and bn are series with...
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