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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.

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Answer:

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 .

[tex]\sum\limits_{n=0}^{\infty} (-1)^n \, b_n[/tex]

The error of the series can be estimated as follows

[tex]R_n = |s-s_n| < b_{n+1}[/tex]

The meaning of the theorem is that   [tex]b_{n+1}[/tex]  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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