## Calculus: Early Transcendentals 8th Edition

Convergant $=\dfrac {25}{3}$
$10-2+0.4-0.08=\sum ^{\infty }_{k=0}10\times \left( -\dfrac {1}{5}\right) ^{k}\Rightarrow \lim _{n\rightarrow \infty }10\times \left( -\dfrac {1}{5}\right) ^{k}=0$ $\sum ^{\infty }_{k=0}\left( 10\times \left( -\dfrac {1}{5}\right) ^{k}\right) -\dfrac {a_{1}}{1-r};a_{1}=10;r=\left( -\dfrac {1}{5}\right) \Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {10}{1-\left( -\dfrac {1}{5}\right) }=\dfrac {25}{3}$