121616949-math.263 - fact than positive term series...

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10.4 Alternating Series 249 Exercises Determine whether each series converges or diverges. 1. X n =1 1 n π/ 4 2. X n =1 n n 2 + 1 3. X n =1 ln n n 2 4. X n =1 1 n 2 + 1 5. X n =1 1 e n 6. X n =1 n e n 7. X n =2 1 n ln n 8. X n =2 1 n (ln n ) 2 9. Find an N so that X n =1 1 n 4 = N X n =1 1 n 4 ± 0 . 005. 10. Find an N so that X n =0 1 e n = N X n =0 1 e n ± 10 - 4 . 11. Find an N so that X n =1 ln n n 2 = N X n =1 ln n n 2 ± 0 . 005. 12. Find an N so that X n =2 1 n (ln n ) 2 = N X n =2 1 n (ln n ) 2 ± 0 . 005. Next we consider series with both positive and negative terms, but in a regular pattern: they alternate, as in the alternating harmonic series for example: X n =1 ( - 1) n - 1 n = 1 1 + - 1 2 + 1 3 + - 1 4 + · · · = 1 1 - 1 2 + 1 3 - 1 4 + · · · . In this series the sizes of the terms decrease, that is, | a n | forms a decreasing sequence, but this is not required in an alternating series. As with positive term series, however, when the terms do have decreasing sizes it is easier to analyze the series, much easier, in
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Unformatted text preview: fact, than positive term series. Consider pictorially what is going on in the alternating harmonic series, shown in figure 10.4 . Because the sizes of the terms a n are decreasing, the partial sums s 1 , s 3 , s 5 , and so on, form a decreasing sequence that is bounded below by s 2 , so this sequence must converge. Likewise, the partial sums s 2 , s 4 , s 6 , and so on, form an increasing sequence that is bounded above by s 1 , so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and...
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