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 see that the series converges for $x \in [-1,1]$ since the terms approach 0 in absolute value and alternate in sign. To find if the series converges at the endpoints of the interval, we can use the alternating series error bound. For $x=1$, the series becomes $\sum_{n=1}^\infty \frac{(-1)^n}{n+3}$ which still satisfies the conditions for the alternating series test, so it converges. Similarly, for $x=-1$, the series becomes $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+3}$ which also converges by the alternating series test. Therefore, 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 $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { n + 3 }$