quiz4mat25

quiz4mat25 - } , either n is even or odd, and in either...

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Quiz 4 Thursday September 4th 1. (5 pts) Assume a n > 0 and b n > 0 for all n . Show that if a n and b n converge, then a n b n converges. Hint: Show that a n b n a n + b n for all n . Solution: see homework 5 solutions. 2. (4 pts) Find the interval of convergence of the power series X n =0 3 n x n ( n + 1) 2 . Don’t forget to check the endpoints. Solution: The radius of convergence is given by R = lim 3 n ( n +1) 2 3 n +1 ( n +2) 2 = 1 3 lim ± n + 2 n + 1 ² 2 = 1 3 . The power series converges at both endpoints because there it’s bounded by the convergent p -series 1 n 2 . Therefore, the interval of convergence is ³ - 1 3 , 1 3 ´ .
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3. In this exercise you’ll prove that the alternating harmonic series converges. Don’t use the alternating series theorem; prove everything from scratch using definitions. (a) (4 pts) Let ( s n ) be any sequence so that the even and odd subsequences both converge to the same limit, i.e. s 2 n s and s 2 n +1 s . Prove s n s . Solution: Let ± > 0 be given. Then N so that 2 n > N , | s 2 n - s | < ± , and M so that 2 n + 1 > M , | s 2 n +1 - s | < ± . Then for n > max { N,M
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Unformatted text preview: } , either n is even or odd, and in either case | s n-s | &lt; . For the rest of the exercise, let s n = 1-1 2 + 1 3- + (-1) n +1 1 n . (b) (5 pts) Prove that ( s 2 n ) is increasing and bounded above. Prove that ( s 2 n +1 ) is decreasing and bounded below. Solution: s 2( n +1)-s 2 n = (-1) 2 n +3 1 2 n + 2 +(-1) 2 n +2 1 2 n + 1 =-1 2 n + 2 + 1 2 n + 1 &gt; so the subsequence is increasing. It is bounded above by 1 because s 2 n = 1-1 2 + 1 3- +(-1) 2 n +1 1 2 n = 1- 1 2-1 3 -- 1 2 n-2-1 2 n-1 -1 2 n , and all terms in the parentheses are positive. Proof that ( s 2 n +1 ) is decreasing and bounded below by 0 is similar. (c) (2 pts) Prove that lim s 2 n = lim s 2 n +1 . Conclude ( s n ) converges. Solution: lim s 2 n = lim s 2 n +1-1 2 n +1 = lim s 2 n +1 because lim 1 2 n +1 = 0....
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quiz4mat25 - } , either n is even or odd, and in either...

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