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hw7_sol - Homework 7 Math 118A Fall 2009 Due on Thursday...

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Homework 7 – Math 118A, Fall 2009 Due on Thursday, December 3rd, 2009 1. Investigate the convergence of a n where (a) a n = n + 1 - n n + 1 . (b) a n = s n + 1 - n n + 1 . Solution: (a) This series converges, because a n = n + 1 - n n + 1 = 1 ( n + 1 + n )( n + 1) c n 3 / 2 , and it follows from the comparison theorem. (b) In this case the series does not converge, because a n = s n + 1 - n n + 1 = 1 q ( n + 1 + n )( n + 1) c n 3 / 4 , and again the comparison theorem proves that it diverges. 2. Let a n and b n converge, with b n > 0 for all n . Suppose that a n /b n L . Prove that lim N →∞ n = N a n n = N b n = L. Solution: Let > 0, then choose N > 0 such that for n N we have | a n /b n - L | < , hence for each n N we can find d n such that | d n | < and a n /b n = L + d n . We then have that n = N a n n = N b n = P n = N Lb n + b n d n P n = N b n (1) = L + P n = N b n d n P n = N b n . (2) 1
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2 We seek to bound the last ratio. n = N b n d n n = N b n P n = N b n | d n | P n = N b n (3) P n = N b n P n = N b n (4) , (5) which gives the desired result. 3. Prove that in R n all norms are equivalent, i.e., given two norms k · k and k · k 0 , there exist positive constants c < d such that c k x k ≤ k x k 0 d k x k . You can prove this by showing that every norm is equivalent to the Euclidean norm, and conclude the result from that. Solution: Let { e i } n i =1 be the canonical basis for R n , and let M = max {k e i k 0 : i = 1 , 2 , . . . , n } . If x R n , then x = n X i =1 x i e i , and therefore k x k 0 n X i =1 | x i |k e i k 0 M n k x k 2 , where the last inequality follows from the Cauchy-Schwarz inequality. This proves the second inequality, but it also proves that the function k · k 0 : ( R n , k · k ) R is continuous. Let B = { x R n |k x k ≤ 1 } . Since B is compact in ( R n , k · k ), we know that k · k 0 achieves its maximum and minimum. Therefore there are constants c > 0 and C > 0 such norm c ≤ k x k 0 C x B.
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