midterm2sol persson fall 09

# X n 1 1 n x n n 3 2 n solution ratio test a n 1 a n x

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the convergence at the end points, for the power series below. X n =1 ( - 1) n x n n · 3 2 n Solution: Ratio test: a n +1 a n = x n +1 ( n + 1) · 9 n +1 · n · 9 n x n = | x | 9 · n n + 1 | x | 9 as n → ∞ | x | 9 < 1 ⇐⇒ | x | < 9 x = 9 : a n = ( - 1) n n = Convergent by the alternating series test x = - 9 : a n = 1 n = Divergent (p-series with p = 1) = I = ( - 9 , 9] 2

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2. (5 points) Show that the series y = X n =0 x 2 n +1 1 · 3 · 5 · · · · · (2 n + 1) is a solution of the differential equation y 0 = 1 + xy. Solution: xy = X n =0 x 2 n +2 1 · 3 · 5 · · · · · (2 n + 1) y 0 = X n =0 (2 n + 1) x 2 n 1 · 3 · 5 · · · · · (2 n + 1) = X n =0 x 2 n 1 · 3 · 5 · · · · · (2 n - 1) = 1 + X n =0 x 2 n +2 1 · 3 · 5 · · · · · (2 n + 1) = y 0 = 1 + xy 3
3. Determine if the series below are absolutely convergent (AC), condition- ally convergent (CC), or divergent (D). a) (5 points) X n =0 2 - 3 sin n 6 n Solution: Check for absolute convergence by bounding the terms: | a n | = 2 - 3 sin n 6 n 5 6 n Compare with the convergent geometric series n =0 ( 5 6 ) n = Series is absolutely convergent (AC). b) (5 points) X n =1 ( - 1) n sin(1 /n 2 ) 1 / 3 Solution: Convergent by the alternating series test, since sin(1 /n 2 ) is decreas- ing for n 1 and sin(1 /n 2 ) sin 0 = 0 as n → ∞ . Since sin(1 /n 2 ) 1 / 3 (1 /n 2 ) 1 / 3 = 1 /n 2 / 3 , use the limit comparison test with b n = 1 /n 2 / 3 to check for absolute convergence: | a n | b n = sin(1 /n 2 ) 1 / 3 1 /n 2 / 3 = 1 n 2 - 1 n 6 · 3! + 1 n 10 · 5!

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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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