Lesson 17a - Telescoping Series

# Example2 determinethesumofthefollowingseries 6 4n 2 9

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Unformatted text preview: s. Example 1 Determine whether the following series converges by producing a telescoping series: x ln x 1 n 1 ln( x) ln( x 1) n 1 S k ln(1) ln(2) ln(2) ln(3) ln(3) ln(4) ln(4) ln(5) ... ln(k ) ln(k 1) S k ln(1) ln(k 1) lim S k lim ln(k 1) k k lim S k k Thus the series diverges! Example 2 Determine the sum of the following series: 6 4n 2 9 n 1 We first must split this up into two or more fractions. To do so, we will use our partial fractioning technique: 6 6 A B 4n 2 9 ( 2n 3)(2n 3) 2n 3 2n 3 6 A(2n 3) B (2n 3) If n = 3/2, we get: 6 A(3 3) A 1 If n=‐3/2 6 B ( 3 3) B 1 6 1 1 4n 2 9 2n 3 2n 3 n 1 n 1 Example 2 Continued 6 1 1 Now we can write the partial sums: 4n 2 9 2n 3 2n 3 n 1 n 1 1 1 1 5 1 1 1 7 1 1 3 9 1 1 5 11 1 1 7 13 ... 1 1 2n 9 2n 3 1 1 2n 7 2n 1 1 1 2n 5 2n 1 1 1 2n 3 2n 3 1 1 1 1 1 1 1 1 1 1 1 1 Sk ... 1 5 1 7 3 9 5 11 7 13 2n 3 2n 3 1 1 1 1 Sk 5 2k 1 2k 1 2k 3 1 1 1 1 lim S k lim k k 5 2k 1 2k 1 2k 3 lim S k k 1 5 Therefore, the sum of the series is 1/5....
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