1 n 2 1 3 1 as n therefore the absolute series is

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- · · · 1 n 2 1 / 3 1 as n → ∞ Therefore, the absolute series is divergent, since n =1 1 /n 2 / 3 is di- vergent (p-series with p = 2 / 3) = Series is conditionally convergent (CC). 4

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4. Find the sum of the series below. a) (5 points) X n =1 2 n ( n + 2) Solution: Write the rational function as a partial fraction: 2 n ( n + 2) = 1 n - 1 n + 2 which gives the partial sum s n = 1 1 - 1 3 + 1 2 - 1 4 + 1 3 - 1 5 + · · · + 1 n - 2 - 1 n + 1 n - 1 - 1 n + 1 + 1 n - 1 n + 2 = 1 + 1 2 - 1 n + 1 - 1 n + 2 3 2 as n → ∞ b) (5 points) X n =0 1 1 + 3 · ( - 1) n n Solution: Split the series: n even : a n = 1 4 n n odd : a n = 1 ( - 2) n = - 1 2 n X n =0 a n = X n =0 1 4 2 n - 1 2 2 n +1 = X n =0 1 16 n - 1 2 · 1 4 n = 1 1 - 1 / 16 - 1 2 · 1 1 - 1 / 4 = 16 15 - 2 3 = 2 5 5
5. (5 points) Find all x that satisfy the equation X n =0 ( - 1) n ( n + 1) x 2 n +2 = 2 9 . Solution: 1 1 - x = X n =0 x n | x | < 1 1 (1 - x ) 2 = X n =1 nx n - 1 = X n =0 ( n + 1) x n | x | < 1 x 2 (1 + x 2 ) 2 = X n =0 ( - 1) n ( n + 1) x 2 n +2 | x 2 | < 1 x 2 (1 + x 2 ) 2 = 2 9 = 9 x 2 = 2(1 + 2 x 2 + x 4 ) = x 4 - 5 2 x 2 + 1 = 0 = x 2 = 5 4 ± r 25 16 - 1 = 5 ± 3 4 = 1 2 , 2 But the series is only convergent for | x 2 | < 1, and therefore x = ± 1 2 6
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• Fall '08
• Reshetiken
• Mathematical Series, G. Melvin T., Melvin T. Wilson, D. Cristofaro-Gardiner E., Cristofaro-Gardiner E. Kim

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