Question

Estimate the magnitude of the error involved in using the sum of the first two terms to approximate the sum of the entire series summation from n equals 1 to infinity left parenthesis negative 1 right parenthesis superscript n plus 1 baseline startfraction left parenthesis 0.08 right parenthesis superscript n over n endfraction

Answer 1

Answer:

the magnitude of the error is 2.9959 × 10⁻⁹

Step-by-step explanation:

By Alternating series estimation theorem

$\sum ^{\infty}_{n=1}(-1)^{n+1}a_n=a_1-a_2+a_3-a_4...+a_n$ is satisfies

(i) $a_n +1\leq a_n$ for all n

(ii) $\lim_{\rightarrow \infty}a_n=0$

The formula for magnitude of error is

$|R_n|=|s-s_n|\leq a_n+1$

Let $a_n = \frac{(0.08)^n}{n}$

We can see $\frac{(0.08)^n^+^1}{n} \leq \frac{(0,08)^n}{n}$

For all n ∈ N

That means $b_{n+1}\leq b_n$ for all   n ∈ N

$\lim_{n \rightarrow \infty}a_n=\lim(\frac{(0.08)^n}{n} )\\\\=\frac{(0.08)^{\infty}}{\infty}$

$\lim_{n\rightarrow \infty}a_n =0$

The alternating series estimation theorem is $|R_s| = |s-s_n|\leq a_{n+1}$

After stop adding the term with n = 6

the series tells that the error is smaller than the term with a

Therefore

$a_7 = \frac{(0.08)^7}{7} \\\\= \frac{20.97152}{7} \times 10^-^9\\\\=2.9959\times10^-^9$

the magnitude of the error is 2.9959 × 10⁻⁹