Solved: "Approximate the sum of the series correct to..."

glamrockqueen7

Answered question

2021-11-08

Approximate the sum of the series correct to four decimal places.

\sum_{n}^{\infty} n = \frac{1 {(- 1)}^{n}}{3^{n} n!}

Answer & Explanation

Mayme

Skilled2021-11-09Added 103 answers

We know that, in an alternating series, the estimate error is smaller than the first neglected term. That's why we will apply. We can see that we have:
$| R_{n} | \leq b_{n + 1} = \frac{1}{3^{n + 1} (n + 1)!} < 0.00001$
Since this can be a tough inequality to solve, we can use graphing calculator to figure out for which n this will aply. We can see that we have:
$\frac{1}{3^{5} 5!} \approx 0.00003 > 0.00001 > \frac{1}{3^{6} 6!} \approx 0.000002$
Therefore, we need to add at least 5 terms for the error to be less than $0.00001$ , that is, for our approximation to be correct to the fourth decimal. Now we have:
$s_{5} = \sum_{k = 1}^{5} \frac{{(- 1)}^{k}}{3^{k} k!} \approx - 0.28347$
The sum of the series is approximately -0.28347

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