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 $\sum_{n = 1}^{\infty}\frac{a}{n^{p}}$ ($p>1$), the remainder $R_N$ (error when using the sum of the first $N$ terms $S_N$ to approximate the sum $S$ of the series) is bounded by $R_N\leq\int_{N}^{\infty}\frac{a}{x^{p}}dx$. Here $a = 5$ and $p=1.2$.

Step2: Evaluate the improper - integral

We have $R_N\leq\int_{N}^{\infty}\frac{5}{x^{1.2}}dx$. First, find the antiderivative of $\frac{5}{x^{1.2}}$. The antiderivative of $x^{-1.2}$ is $\frac{x^{-1.2 + 1}}{-1.2+1}=-5x^{-0.2}$, so $\int_{N}^{\infty}\frac{5}{x^{1.2}}dx=\lim_{b\rightarrow\infty}\int_{N}^{b}\frac{5}{x^{1.2}}dx=\lim_{b\rightarrow\infty}\left[- 5x^{-0.2}\right]{N}^{b}=\lim{b\rightarrow\infty}\left(-\frac{5}{b^{0.2}}+\frac{5}{N^{0.2}}\right)=\frac{5}{N^{0.2}}$.

Step3: Set up the error inequality

We want $R_N\leq0.000001$. So, $\frac{5}{N^{0.2}}\leq0.000001$.

Step4: Solve the inequality for N

Cross - multiply to get $5\leq0.000001N^{0.2}$. Then $N^{0.2}\geq\frac{5}{0.000001}=5\times10^{6}$. Raise both sides to the power of 5 (since $0.2=\frac{1}{5}$) to obtain $N\geq(5\times10^{6})^{5}=5^{5}\times10^{30}=3125\times10^{30}=3.125\times10^{33}$. We can write $N\approx10^{33}$.

Answer:

$33$