## University Calculus: Early Transcendentals (3rd Edition)

Write the series as a sum of two series $\Sigma_{n=1}^{\infty}(1-\dfrac{7}{4^n})=\Sigma_{n=1}^{\infty}1-\Sigma_{n=1}^{\infty}7(\dfrac{1}{4^n})$ Here, $-\Sigma_{n=1}^{\infty}7(\dfrac{1}{4^n})$ converges because it a geometric series with common ratio $r=\dfrac{1}{4}\lt 1$ But $\Sigma_{n=1}^{\infty}1$ diverges by the n-th term test. Thus, the final series will be divergent.