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Unformatted text preview: Math 138 Assignment 7 Solutions 1. Determine whether or not the following series are convergent. If so, find the sum. If not, explain why. (a) ∞ X n =1 1 + 2 n 4 n 1 ∞ X n =1 1 + 2 n 4 n 1 = ∞ X n =1 1 4 n 1 + ∞ X n =1 2 n 4 n 1 = ∞ X n =1 4 1 4 n + ∞ X n =1 4 2 4 n These are both geometric series, the first with a = 1 and r = 1 4 , the second with a = 2, r = 1 2 . Therefore, ∞ X n =1 1 + 2 n 4 n 1 = 1 1 1 4 + 2 1 1 2 = 4 3 + 4 = 16 3 (b) ∞ X n =1 1 n ( n + 2) (this is a telescoping series) By partial fractions, 1 n ( n + 2) = 1 2 1 n 1 n + 2 Then the partial sum s n is: s n = 1 2 1 1 3 + 1 2 1 4 + 1 3 1 5 + 1 4 1 6 + ... + 1 n 2 1 n + 1 n 1 1 n + 1 + 1 n 1 n + 2 Most terms cancel and all that is left is: s n = 1 2 1 + 1 2 1 n + 1 1 n + 2 → 3 4 as n → ∞ Thus ∞ X n =1 1 n ( n + 2) = 3 4 (c) ∞ X n =1 e 1 /n The series diverges by the Test for Divergence, since a n = e 1 /n → 1 as n → ∞ . (d) We find the n th partial sum to show convergence. The integral test would also work (to determine convergence but would not yield the sum). To find the sum we decompose as 1 n ( n + 1)( n 1) = 1 2( n + 1) + 1 2( n 1) 1 n . Then we write the first few terms as follows ∞ X n =2 1 2( n + 1) + 1 2( n 1) 1 n = 1 6 + 1 2 1 2 + 1 8 + 1 4 1 3 + 1 10 + 1 6 1 4 + 1 12 + 1 8 1 5 + ··· In this case it’s not as obvious as to what might cancel. One thing for certain is that we can cancel the 1 2 terms immediately. We might also be tempting is to cancel the 1 4 terms but consider instead the following grouping (based on matching colours)...
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 Fall '07
 Anoymous
 Math, Calculus, lim

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