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Please help with following questions: The Alternating Series Test. If an alternating series E(-1)than = a1 - az tag - ay + as ... where

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The Alternating Series Test. If an alternating series E(-1)"than = a1 - az tag - ay + as ... where an, > 0 for all n satisfies . Ontl on for all n large (i.c., the sequence {on ) is eventually decreasing). lim on =0 then the series is convergent. 1. _(-1)"-1 3 An - 1 2. iM 8 13 + 1 Approximation of Alternating Series If an alternating series >(-1)"+lan satisfies the hypotheses of the alternating series, and if S is the sum of the N= 1 series, then (a) 5 lies between any two consecutive partial sums, that is depending on which partial sum is larger. (b) If S is approximated by s,, then The size of the error is no larger than the absolute value of the first neglected term.4. Consider the series (-1)-1 1=1 n! (a) Show that the series converges and find the sum of the series correct to two decimal places. =0.5 1 =0.GGG =0.625 = 0.6333 - 0.6319 (b) How many term of the series do we nood to add in order to find the sum with absolute error 10000 This shows n + 1 2 9, or n > 8. That means we need to add 8 terms. 5. Consider the series n4 (a) Show that the series converges and find the sum of the series correct to two decimal places. (b) How many term of the series do we need to add in order to find the sum with absolute error lo,| is convergent. Note that if ) on is a series with positive terms, then Ja,| = a, and so absolute convergence is the same as convergence in this case. . A series ) on is called conditionally convergent if it is convergent but not absolutely convergent. Theorem. If a series _ an is absolutely convergent, then it is convergent. Ratio Test for Absolutely Convergence Let ) an be a series with non-zero terms (i.e, a, # 0 for all n). Suppose that lim P . If p 1 then the series diverges. . If p = 1 the test is inconclusive. Determine whether the series is absolutely convergent, conditionally convergent, or divergent. (-1 )(at1)n 1. n + 1 2. T(-1)" Vn -2) m 3. 4. (-1)-1_ 72 + 4 iM: in 5. (-1) " In(n + 1)

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