Find the interval of convergence of the series. $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { n + 3 }$
A) $[ - 1,1 ]$
B) $( - 1,1 )$
C) $( - 1,1 ]$
D) diverges everywhere
E) $[ - 1,1 )$

Question

Quizplus · Accepted Answer

We can use the alternating series test to show that the series converges for $x=-1$ and $x=1$. To find the interval of convergence, we can use the ratio test: $$\lim_{n	o\infty}\left|\frac{\frac{(-1)^{n+1}x^{n+1}}{n+4}}{\frac{(-1)^nx^n}{n+3}}ight|=\lim_{n	o\infty}\left|\frac{x}{n+4}ight|=0<1$$ Therefore, the series converges for all $x$ values except $x=-4$. This means the interval of convergence is $(-1,1]$.

Find the Interval of Convergence of the Series ∑n=1∞(−1)nxnn+3\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { n + 3 }∑n=1∞​n+3(−1)nxn​

Find the Interval of Convergence of the Series $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { n + 3 }$