use the sum of the first 10 terms to approximate the sum of the series. (round your answer to five decimal…

Explanation:

Step1: Integrate the function

Recall the integral formula (\int a^{x}dx=\frac{a^{x}}{\ln a}+C) ((a > 0,a\neq1)). For the integral (\int\frac{1}{3^{x}}dx=\int3^{-x}dx), let (u=-x), then (du=-dx). So (\int3^{-x}dx=-\frac{3^{-x}}{\ln 3}+C).

Step2: Evaluate the definite - integral

Using the fundamental theorem of calculus (\int_{10}^{\infty}\frac{1}{3^{x}}dx=\lim_{t\rightarrow\infty}\int_{10}^{t}3^{-x}dx). [ \begin{align*} \lim_{t\rightarrow\infty}\int_{10}^{t}3^{-x}dx&=\lim_{t\rightarrow\infty}\left[-\frac{3^{-x}}{\ln 3}\right]{10}^{t}\ &=\lim{t\rightarrow\infty}\left(-\frac{3^{-t}}{\ln 3}+\frac{3^{-10}}{\ln 3}\right) \end{align*} ] Since (\lim_{t\rightarrow\infty}3^{-t}=\lim_{t\rightarrow\infty}\frac{1}{3^{t}} = 0).

Answer:

(\frac{1}{3^{10}\ln 3}\approx\frac{1}{59049\times1.098612289}\approx0.00001543)