Find the sum of the series [infinity] (−1)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! ≤ 1 n → 1 so 1 n! → 1 as n → [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! − 1 1! + 1 2! − 1 3! + 1 4! − 1 5! + 1 6! − 1 7! + ⋯ = 1 − 1 + 1 2 − 1 6 + 1 24 − 1 120 + 1 − 1 5040 + ⋯ Notice that b7 = 1 5040 < 1 5000 = and s6 = 1 − 1 + 1 2 − 1 6 + 1 24 − 1 120 + 1 720 ≈ (rounded to six decimal places). By the Alternating Series Estimation Theorem we know that |s − s6| ≤ b7 ≤ 0.0002. This error of less than 0.0002 does not affect the third decimal place, so we have s ≈ 0.368 correct to three decimal places.