$\text { Identify the interval of convergence of a power series } \sum _ { n = 1 } ^ { \infty } ( - 9 ) ^ { n } x ^ { 2 n } \text {. }$
A) $- \frac { 1 } { 3 }

Question

$\text { Identify the interval of convergence of a power series } \sum _ { n = 1 } ^ { \infty } ( - 9 ) ^ { n } x ^ { 2 n } \text {. }$ 
A) $- \frac { 1 } { 3 } < x < \frac { 1 } { 3 }$
B) $0 < x < \frac { 1 } { 3 }$
C) $- \frac { 1 } { 9 } < x < \frac { 1 } { 9 }$
D) $- \frac { 1 } { 9 } < x < 0$
E) $0 < x < \frac { 1 } { 9 }$

Quizplus · Accepted Answer

The interval of convergence can be found using the Ratio Test. The series is $\sum_{n=1}^{\infty} (-9)^n x^{2n}$. Let $a_n = (-9)^n x^{2n}$. The Ratio Test gives $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-9)^{n+1} x^{2(n+1)}}{(-9)^n x^{2n}} \right| = \left| -9x^2 \right| < 1$. Solving for $x$, we get $-\frac{1}{3} < x < \frac{1}{3}$.

Identify the interval of convergence of a power series ∑n=1∞(−9)nx2n. \text { Identify the interval of convergence of a power series } \sum _ { n = 1 } ^ { \infty } ( - 9 ) ^ { n } x ^ { 2 n } \text {. } Identify the interval of convergence of a power series ∑n=1∞​(−9)nx2n.

$\text { Identify the interval of convergence of a power series } \sum _ { n = 1 } ^ { \infty } ( - 9 ) ^ { n } x ^ { 2 n } \text {. }$