## Calculus: Early Transcendentals 8th Edition

(a) Since it is given that the series converges when $x=-4$, this implies that the series converges for all $x$ which lie in the interval $(-4,4)$ and $\Sigma_{n=0}^{\infty}c_{n}$ is obtained after substituting $x=1$ in $\Sigma_{n=0}^{\infty}c_{n}x^{n}$ Hence, the given series is convergent. (b) Since it is given that the series converges when $x=6$, this implies that the interval of convergence is smaller and equal to $[-6,6)$ and $\Sigma_{n=0}^{\infty}c_{n}8^{n}$ is obtained after substituting $x=8$ in $\Sigma_{n=0}^{\infty}c_{n}x^{n}$ Hence, the given series is divergent. (c) Since it is given that the series converges when $x=-4$, this implies that the series converges for all $x$ which lie in the interval $(-4,4)$ and $\Sigma_{n=0}^{\infty}c_{n}(-3)^{n}$ is obtained after substituting $x=-3$ in $\Sigma_{n=0}^{\infty}c_{n}x^{n}$ Hence, the given series is convergent. (d) Since it is given that the series converges when $x=6$, this implies that the interval of convergence is smaller and equal to $[-6,6)$ and $\Sigma_{n=0}^{\infty}(-1)^{n}c_{n}9^{n}$ is obtained after substituting $x=-9$ in $\Sigma_{n=0}^{\infty}c_{n}x^{n}$ Hence, the given series is divergent.