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Mathematics, 24.07.2020 23:01 1969976289
EXAMPLE 4 Find the sum of the series [infinity] (β1)n n! n = 0 correct to three decimal places. SOLUTION We first observe that the series is convergent by the Alternating Series Test because (i) 1 (n + 1)! = 1 n!(n + 1) 1 n! (ii) 0 < 1 n! β€ 1 n β so 1 n! β as n β [infinity]. To get a feel for how many terms we need to use in our approximation, let's write out the first few terms of the series: s = 1 0! β 1 1! + 1 2! β 1 3! + 1 4! β 1 5! + 1 6! β 1 7! + β― = 1 β 1 + 1 2 β 1 6 + 1 24 β 1 120 + 1 β 1 5040 + β― Notice that b7 = 1 5040 < 1 5000 = and s6 = 1 β 1 + 1 2 β 1 6 + 1 24 β 1 120 + 1 720 β (rounded to six decimal places). By the Alternating Series Estimation Theorem we know that |s β s6| β€ b7 β€ 0.0002. This error of less than 0.0002 does not affect the third decimal place, so we have s β 0.368 correct to three decimal places. Need Help? Read It
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EXAMPLE 4 Find the sum of the series [infinity] (β1)n n! n = 0 correct to three decimal places. SOLU...
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