fall17mth143.practice2.2-InfiniteSeries.pdf

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n = 1 n n 2 + 1 Answer: Answer(s) submitted: (incorrect) Correct Answers: div 17. (1 point) Consider the series n = 1 n 2 n + 8 Determine whether the series converges, and if it converges, determine its value. Converges (y/n): Value if convergent (blank otherwise): Answer(s) submitted: (score 0.5) Correct Answers: n 18. (1 point) Consider the sequence given by x n = ln ( 3 n + 3 ) - ln ( 3 n ) . (i) Compute lim n x n . lim n x n = (ii) Now compute N n = 1 x n in terms of N . N n = 1 x n = (iii) Finally, use your answer to part (ii) to compute n = 1 x n . n = 1 x n = If the limit or series is , write ”Infinity” If the limit or series is - , write ”-Infinity” If the limit does not exist, write ”DNE”. If the series diverges but not to or - , write ”DIV”. Solution: Solution: The infinite sum n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) is the limit of the partial sums, lim n s n , where s k = k n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) . We can easily see by induction that the partial sums are s k = 4

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ln ( 2 ( k + 1 )) - ln ( 2 ) , a sequence which diverges: s 1 = 1 n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) = ln ( 2 ( 1 + 1 )) - ln ( 2 · 1 ) = ln ( 4 ) - ln ( 2 ) s 2 = 2 n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) = 1 n = 1 ( ln ( 2 ( n + 1 )) - ln ( 2 n ))+ 2 n = 2 ( ln ( 2 ( n + 1 )) - ln ( 2 n )) = s 1 + ln ( 2 ( 2 + 1 )) - ln ( 2 · 2 ) = ln ( 4 ) - ln ( 2 )+ ln ( 6 ) - ln ( 4 ) = ln ( 6 ) - ln ( 2 ) ··· s k = k n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) = k - 1 n = 1 ( ln ( 2 ( n + 1 )) - ln ( 2 n ))+ k n = k ( ln ( 2 ( n + 1 )) - ln ( 2 n )) = s k - 1 + ln ( 2 ( k + 1 )) - ln ( 2 k ) = ln ( 2 ( k - 1 + 1 )) - ln ( 2 )+ ln ( 2 ( k + 1 )) - ln ( 2 k ) = ln ( 2 k ) - ln ( 2 )+ ln ( 2 ( k + 1 )) - ln ( 2 k ) = ln ( 2 ( k + 1 )) - ln ( 2 ) However, we could also have solved this by the Limit Com- parison Test, because n = 1 ln ( 2 ( n + 1 )) - ln ( 2 n ) = n = 1 ln 2 ( n + 1 ) 2 n = n = 1 ln n + 1 n = n = 1 ln 1 + 1 n So comparing it with the series n = 1 1 n gives a favorable result: lim n ln ( 1 + 1 n ) 1 n = lim x 0 ln ( 1 + x ) x = lim x 0 ( 1 1 + x ) 1 by l’Hospital’s rule = lim x 0 1 1 + x = 1 Thus we’ve bound together the behavior of the given sequence with the sequence n = 1 1 n , which diverges. So either of the above two methods could show the diver- gence of this series.
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