Explanation If { a n } is a sequence of real numbers. Then the expression S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is termed as sequence of n th partial sum, i.e., the sum of first n terms of a sequence and S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is known as sum of infinite terms of sequence termed as infinite series. The given series is termed as positive term series if in the series S = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n , a n > 0 ∀ n ∈ N . An infinite series S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is said to be convergent or divergent if its n th partial sum S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is convergent to finite value or divergent to infinity. Necessary and sufficient condition for convergence Necessary and sufficient condition that infinite series S = ∑ n = 1 ∞ a n is said to be convergent if lim n → ∞ a n = 0 . Suppose a n ≥ 0 ∀ n ∈ N , then S = ∑ n = 1 ∞ a n is said to be convergentif and only if S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is bounded above. Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Comparison test: Suppose 0 ≤ a n ≤ b n ∀ n ≥ k for some k. Then (1) If ∑ n = 1 ∞ b n is convergent then ∑ n = 1 ∞ a n is also convergent (2) If ∑ n = 1 ∞ a n is divergentthen ∑ n = 1 ∞ b n is alsodivergent Limit Comparison Test: Suppose a n , b n ≥ 0 ∀ n ∈ N and lim n → ∞ a n b n = L (1) If L ∈ R , L > 0 , then both ∑ n = 1 ∞ a n and ∑ n = 1 ∞ b n converge or diverge together. (2) If L ∈ R , L = 0 and ∑ n = 1 ∞ b n converges, then ∑ n = 1 ∞ a n also converges (3) If L → ∞ and ∑ n = 1 ∞ b n diverges, then ∑ n = 1 ∞ a n also diverges Cauchy Test or Cauchy condensation Test : Suppose a n ≥ 0 ∀ n ∈ N and a n + 1 ≤ a n , then ∑ n = 1 ∞ a n converges if and only if ∑ k = 0 ∞ 2 k a 2 k converges. p- Series Test : (1) ∑ n = 1 ∞ 1 n p converges if p > 1 and diverges if p ≤ 1 (2) ∑ n = 2 ∞ 1 n ( log n ) p converges if p > 1 and diverges if p ≤ 1 Ratio Test: D’ Alembert Test: Suppose a n > 0 and lim n → ∞ a n a n + 1 = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges. (2) If L ∈ R , L < 1 , then ∑ n = 1 ∞ a n diverges. (3) If L ∈ R , L = 1 , then test gets fail Raabe’s Test: Suppose a n > 0 and lim n → ∞ n ( a n a n + 1 − 1 ) = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges...

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 10.4, Problem 33E

To determine

The convergence of series using partial sums and whether the given infinite series converges absolutely.

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Chapter 10.4, Problem 32E

Chapter 10.4, Problem 34E

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Total marks 15 4. : Let f R2 R be defined by f(x1, x2) = 2x²- 8x1x2+4x+2. Find all local minima of f on R². [10 Marks] (ii) Give an example of a function f R2 R which is neither bounded below nor bounded above, and has no critical point. Justify briefly your answer. [5 Marks]

4. Let F RNR be a mapping. (i) x ЄRN ? (ii) : What does it mean to say that F is differentiable at a point [1 Mark] In Theorem 5.4 in the Lecture Notes we proved that if F is differentiable at a point x E RN then F is continuous at x. Proof. Let (n) CRN be a sequence such that xn → x ЄERN as n → ∞. We want to show that F(xn) F(x), which means F is continuous at x. Denote hnxn - x, so that ||hn|| 0. Thus we find ||F(xn) − F(x)|| = ||F(x + hn) − F(x)|| * ||DF (x)hn + R(hn) || (**) ||DF(x)hn||+||R(hn)||| → 0, because the linear mapping DF(x) is continuous and for all large nЄ N, (***) ||R(hn) || ||R(hn) || ≤ → 0. ||hn|| (a) Explain in details why ||hn|| → 0. [3 Marks] (b) Explain the steps labelled (*), (**), (***). [6 Marks]

4. In Theorem 5.4 in the Lecture Notes we proved that if F: RN → Rm is differentiable at x = RN then F is continuous at x. Proof. Let (xn) CRN be a sequence such that x → x Є RN as n → ∞. We want F(x), which means F is continuous at x. to show that F(xn) Denote hn xnx, so that ||hn||| 0. Thus we find ||F (xn) − F(x) || (*) ||F(x + hn) − F(x)|| = ||DF(x)hn + R(hn)|| (**) ||DF(x)hn|| + ||R(hn) || → 0, because the linear mapping DF(x) is continuous and for all large n = N, |||R(hn) || ≤ (***) ||R(hn)|| ||hn|| → 0. Explain the steps labelled (*), (**), (***) [6 Marks] (ii) Give an example of a function F: RR such that F is contin- Total marks 10 uous at x=0 but F is not differentiable at at x = 0. [4 Marks]

Chapter 10 Solutions

Calculus: Early Transcendentals

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 5PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E

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