## Calculus: Early Transcendentals 8th Edition

$3-4+\dfrac {16}{3}-\dfrac {64}{9}=\sum ^{\infty }_{k=0}3\times \left( -\dfrac {4}{3}\right) ^{k}\Rightarrow \lim _{k\rightarrow \infty }3\times \left( -\dfrac {4}{3}\right) ^{k}\neq 0$ Then the series diverges