## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.6 Absolute Convergence and the Radio and Root Tests - 11.6 Exercises - Page 784: 48

#### Answer

$s_{10}\approx 1.98828$, error $\leq 0.012$

#### Work Step by Step

By the power of calculator $s_{10}=\Sigma_{n=1}^{10}\frac{n}{2^{n}}\approx 1.98828$ $R_{n}=\dfrac{a_{n+1}}{a_{n}}=\dfrac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^{n}}}=\frac{1}{2}+\frac{1}{2n}$ $\lim\limits_{n \to \infty}(\frac{1}{2}+\frac{1}{2n})=\frac{1}{2}$ Now, $R_{10}\leq \frac{a_{11}}{1-r_{11}}=\dfrac{a_{11}}{1-\dfrac{a_{12}}{a_{11}}}$ error $\leq 0.012$ Hence, $s_{10}\approx 1.98828$, error $\leq 0.012$

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