21d_00w_1s

21d_00w_1s - = lim n ( n + 1) 2 | x + 3 | 2 n 2 = | x + 3 |...

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Bender Math 21D Outline of Solutions to First Midterm Feb. 3, 2000 1. Often more than one test can be used, so several alternative solutions are be given. (a) Diverges . Limit comparison test with a n =1 /n 1 / 2 , a divergent p -series. (lim n →∞ b n /a n =1) Also, integral test: R ( x +3) - 1 / 2 dx =2( x +3) 1 / 2 + C . (b) Converges . Alternating series test: signs alternate, terms go to zero, absolute values of terms decrease. (c) Diverges . Limit comparison test with a n =1 /n , a divergent p -series. (lim n →∞ b n /a n =1) Also, rewrite terms as 2 - n +1 /n . Since 2 - n converges (geometric series) and 1 /n diverges ( p -series), the sum of the two diverges. (d) Diverges . The terms do not go to zero. (e) Diverges . Ratio test: a n +1 /a n =6 2 / 3 3 =36 / 27 > 1. Also root test, or geometric series ( r =6 2 / 3 3 > 1), or the terms do not go to zero. (f) Converges . Same reasoning as (e): The ratio tests give L =27 / 36 and the geometric series has r =27 / 36. (Terms go to zero, so the last test mentioned in (e) cannot be used.) 2. Use the root test or ratio test to find R : L = lim n →∞ a n +1 a n
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Unformatted text preview: = lim n ( n + 1) 2 | x + 3 | 2 n 2 = | x + 3 | 2 . This gives convergence for | x + 3 | < 2 and divergence for | x + 3 | > 2. Thus R = 2. (Alternatively, you can do what I mentioned in class: Replace x + 3 with R and set L = 1: 1 = L = lim = R/ 2 and so R = 2.) The endpoints of the interval are given by x + 3 = 2 and x + 3 =-2. The former gives the series n 2 and the latter gives n 2 (-1) n . Both diverge since the terms do not go to zero. Thus the interval of convergence is-5 < x <-1. 3. Since e x = x n /n !, replacing x by-2 x 2 gives e-2 x 2 = X n =0 (-2 x 2 ) n n ! = X n =0 (-2) n x 2 n n ! . Multiply by 1 + x : (1 + x ) e-2 x 2 = X n =0 (-2) n ( x 2 n + x 2 n +1 ) n ! . The coecient of x 10 comes from the n = 5 term in the sum. The coecient is (-2) 5 / 5! =-4 / 15. This is also the coecient of x 11 ....
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