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

use the sum of the first 10 terms to approximate the sum of the series. (round your answer to five decimal places.)\n sum _ { n = 1 } ^ { infty } 3 ^ { - n } cos ^ { 2 } ( n ) \nestimate the error. (round your answer to eight decimal places.)\n r _ { 10 } leq int _ { 10 } ^ { infty } \frac { 1 } { 3 ^ { x } } d x = \n

use the sum of the first 10 terms to approximate the sum of the series. (round your answer to five decimal places.)\n sum _ { n = 1 } ^ { infty } 3 ^ { - n } cos ^ { 2 } ( n ) \nestimate the error. (round your answer to eight decimal places.)\n r _ { 10 } leq int _ { 10 } ^ { infty } \frac { 1 } { 3 ^ { x } } d x = \n

Answer

Explanation:

Step1: Integrate the function

We know that (\int\frac{1}{3^{x}}dx=\int3^{-x}dx). Using the formula (\int a^{u}du=\frac{a^{u}}{\ln a}+C) (here (a = 3), (u=-x), (du=-dx)), we have (\int3^{-x}dx=-\frac{3^{-x}}{\ln 3}+C).

Step2: Evaluate the definite - integral

[ \begin{align*} \int_{10}^{\infty}\frac{1}{3^{x}}dx&=\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}\approx0.00000595)