## Multivariable Calculus, 7th Edition

The series is convergent and its sum is $\frac{32}{7}$.
$\Sigma^{\infty}_{n=1}[(0.8)^{n-1} - (0.3)^{n}]$ For $\Sigma^{\infty}_{n=1}(0.8)^{n-1}$, $a=1$ and $r=-0.8$ Since $|r|= 0.8 \lt 1$, therefore $\Sigma^{\infty}_{n=1}(0.8)^{n-1}$ is convergent and its sum is $=\frac{a}{1-r}$ $=\frac{1}{1-0.8}$ $=\frac{1}{0.2}$ $=5$ For $\Sigma^{\infty}_{n=1}(0.3)^{n} = \Sigma^{\infty}_{n=1}(0.3)(0.3)^{n-1}$, $a=0.3$ and $r=0.3$ Since $|r|=0.3 \lt 1$, the series $\Sigma^{\infty}_{n=1}(0.3)^{n}$ converges and its sum is $=\frac{a}{1-r}$ $=\frac{0.3}{1-0.3}$ $=\frac{3}{7}$ Since both sums are convergent, the series is convergent and its sum is $5-\frac{3}{7} = \frac{32}{7}$