how many terms of the convergent series ∑(n = 1 to ∞) 5 / n^1.2 should be used to estimate its value with…

Explanation:

Step1: Recall the remainder formula for p - series

For a p - series $\sum_{n = 1}^{\infty}\frac{a}{n^{p}}$ ($p>1$, $a>0$), the remainder $R_N=\sum_{n = N+1}^{\infty}\frac{a}{n^{p}}\leq\int_{N}^{\infty}\frac{a}{x^{p}}dx$. Here $a = 5$, $p=1.2$.

Step2: Calculate the integral

We have $\int_{N}^{\infty}\frac{5}{x^{1.2}}dx=\lim_{b\rightarrow\infty}\int_{N}^{b}5x^{- 1.2}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $\lim_{b\rightarrow\infty}\left[5\frac{x^{-1.2 + 1}}{-1.2+1}\right]{N}^{b}=\lim{b\rightarrow\infty}\left[\frac{5x^{-0.2}}{-0.2}\right]{N}^{b}=\lim{b\rightarrow\infty}\left(-25x^{-0.2}\right)\big|{N}^{b}$. Evaluating the limit: $\lim{b\rightarrow\infty}\left(-25b^{-0.2}+25N^{-0.2}\right)=25N^{-0.2}$.

Step3: Set up the error inequality

We want $R_N\leq0.000001$. So, $25N^{-0.2}\leq0.000001$.

Step4: Solve the inequality for N

First, rewrite the inequality as $\frac{25}{N^{0.2}}\leq0.000001$. Then, cross - multiply to get $25\leq0.000001N^{0.2}$. Next, divide both sides by $0.000001$: $N^{0.2}\geq\frac{25}{0.000001}=25000000$. Raise both sides to the power of 5 (since $0.2=\frac{1}{5}$) to solve for $N$: $N\geq(25000000)^{5}=9.765625\times10^{35}\approx10^{36}$.

Answer:

$36$