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.

In this case, if you sum the first seven terms of your sum, you get 0.0002, that means that your error is less than 0.00002 , in other words, if you sum 6 terms of your sum, 4 terms of your result are correct, (because the error is less than 0.00002).

Step-by-step explanation:

Remember what the alternating series theorem says, basically it states that for a convergent alternating series .

The error of the series can be estimated as follows

The meaning of the theorem is that    is an upper bound of the n-error of your sum.

In this case, if you sum the first seven terms of your sum, you get 0.0002, that means that your error is less than 0.00002 , in other words, if you sum 6 terms of your sum, 4 terms of your result are correct, (because the error is less than 0.00002).

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step-by-step explanation:

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