how many terms of the convergent series ∑(n = 1 to ∞) 8 / n^1.1 should be used to estimate its value with…

Explanation:

Step1: Recall the remainder formula for p - series

For a p - series $\sum_{n = 1}^{\infty}\frac{a}{n^{p}}$ ($p>1$), the remainder $R_N$ (the error when using the sum of the first $N$ terms $S_N$ to approximate the sum $S$ of the series) is bounded by the integral. That is, $R_N=\sum_{n = N+1}^{\infty}\frac{a}{n^{p}}\leq\int_{N}^{\infty}\frac{a}{x^{p}}dx$. Here, $a = 8$ and $p=1.1$.

Step2: Evaluate the improper - integral

First, evaluate $\int_{N}^{\infty}\frac{8}{x^{1.1}}dx$. Using the power - rule for integration $\int x^{r}dx=\frac{x^{r + 1}}{r+1}+C$ ($r\neq - 1$), we have $\int_{N}^{\infty}\frac{8}{x^{1.1}}dx=\lim_{b\rightarrow\infty}\int_{N}^{b}8x^{-1.1}dx$. [ \begin{align*} \lim_{b\rightarrow\infty}\int_{N}^{b}8x^{-1.1}dx&=\lim_{b\rightarrow\infty}\left[8\times\frac{x^{-1.1 + 1}}{-1.1+1}\right]{N}^{b}\ &=\lim{b\rightarrow\infty}\left[8\times\frac{x^{-0.1}}{-0.1}\right]{N}^{b}\ &=\lim{b\rightarrow\infty}\left[-80x^{-0.1}\right]{N}^{b}\ &=\lim{b\rightarrow\infty}\left(-\frac{80}{b^{0.1}}+\frac{80}{N^{0.1}}\right) \end{align*} ] As $b\rightarrow\infty$, $\lim_{b\rightarrow\infty}\frac{-80}{b^{0.1}} = 0$. So, $\int_{N}^{\infty}\frac{8}{x^{1.1}}dx=\frac{80}{N^{0.1}}$.

Step3: Set up the error inequality

We want $R_N\leq0.00001$. Since $R_N\leq\int_{N}^{\infty}\frac{8}{x^{1.1}}dx$, we set up the inequality $\frac{80}{N^{0.1}}\leq0.00001$.

Step4: Solve the inequality for N

Cross - multiply the inequality $\frac{80}{N^{0.1}}\leq0.00001$ to get $80\leq0.00001N^{0.1}$. Then, divide both sides by $0.00001$: $N^{0.1}\geq\frac{80}{0.00001}=8000000$. Raise both sides to the power of 10 to solve for $N$: $N\geq(8000000)^{10}=8^{10}\times10^{60}\approx1.073741824\times10^{7}$. So, $N\approx10^{7}$.

Answer:

$7$