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Unformatted text preview: Numerical Sequences and Series Written by MenGen Tsai email: b89902089@ntu.edu.tw 1. Prove that the convergence of { s n } implies convergence of { s n } . Is the converse true? Solution: Since { s n } is convergent, for any > 0, there exists N such that  s n s  < whenever n ≥ N . By Exercise 1.13 I know that  s n  s  ≤  s n s  . Thus,  s n  s  < , that is, { s n } is convergent. The converse is not true. Consider s n = ( 1) n . 2. Calculate lim n →∞ ( √ n 2 + n n ). Solution: √ n 2 + n n = n √ n 2 + n + n = 1 q 1 /n + 1 + 1 → 1 2 as n → ∞ . 3. If s n = √ 2 and s n +1 = q 2 + √ s n ( n = 1 , 2 , 3 ,... ) , prove that { s n } converges, and that s n < 2 for n = 1 , 2 , 3 ,... . Proof: First, I show that { s n } is strictly increasing. It is trivial that s 2 = q 2 + √ s 1 = r 2 + q √ 2 > √ 2 = s 1 . Suppose s k > s k 1 when 1 k < n . By the induction hypothesis, s n = q 2 + √ s n 1 > q 2 + √ s n 2 = s n 1 By the induction, { s n } is strictly increasing. Next, I show that { s n } is bounded by 2. Similarly, I apply the induction again. Hence { s n } is strictly increasing and bounded, that is, { s n } converges. 4. 5. 6. 7. Prove that the convergence of ∑ a n implies the convergence of X √ a n n if a n ≥ 0. Proof: By Cauchy’s inequality, k X n =1 a n k X n =1 1 n 2 ≥ k X n =1 a n √ a n n for all n ∈ N . Also, both ∑ a n and ∑ 1 n 2 are convergent; thus ∑ k n =1 a n √ a n n is bounded. Besides, √ a n n ≥ 0 for all n . Hence ∑ √ a n n is convergent. 8. 9. Find the radius of convergence of each of the following power series: ( a ) X n 3 z n , ( b ) X 2 n n ! z n , ( c ) X 2 n n 2 z n , ( d ) X n 3 3 n z n . 2 Solution: (a) α n = ( n 3 ) 1 /n → 1 as n → ∞ . Hence R = 1 /α = 1. (b) α n = (2 n /n !) 1 /n = 2 / ( n !) 1 /n → 0 as n → ∞ . Hence R = + ∞ . (c) α n = (2 n /n 2 ) 1 /n → 2 / 1 = 2 as n → ∞ . Hence R = 1 /α = 1 / 2. (d) α n = ( n 3 / 3 n ) 1 /n → 1 / 3 as n → ∞ . Hence R = 1 /α = 3. 10. 11. Suppose a n > 0, s n = a 1 + ... + a n , and ∑ a n diverges. (a) Prove that ∑ a n 1+ a n diverges. (b) Prove that a N +1 s N +1 + ... + a N + k s N + k ≥ 1 s N s N + k and deduce that ∑ a n s n diverges. (c) Prove that a n s 2 n ≤ 1 s n 1 1 s n and deduce that ∑ a n s 2 n converges. (d) What can be said about X a n 1 + na n and X a n 1 + n 2 a n ? Proof of (a): Note that a n 1 + a n → ⇔ 1 1 a n + 1 → ⇔ 1 a n → ∞ ⇔ a n → as n → ∞ . If ∑ a n 1+ a n converges, then a n → 0 as n → ∞ . Thus for some = 1 there is an N 1 such that a n < 1 whenever n ≥ N 1 . Since ∑ a n 1+ a n converges, for any > 0 there is an N 2 such that a m 1 + a m + ... + a n 1 + a n < 3 all n > m ≥ N 2 . Take N = max( N 1 ,N 2 ). Thus > a m 1 + a m + ... + a n 1 + a n > a m 1 + 1 + ... + a n 1 + 1 = a m + ... + a n 2 for all n > m ≥ N . Thus a m + ... + a n < 2 for all n > m ≥ N . It is a contradiction. Hence ∑ a n...
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This note was uploaded on 11/07/2009 for the course MATH 410 taught by Professor Ddd during the Spring '09 term at Punjab Engineering College.
 Spring '09
 ddd
 Sequences And Series

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