The interval of convergence of the power series $\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { n ^ { 2 } + 3 }$ is

A)(-1,1)
B)[-1,1)
C)(-1,1]
D)[-1,1]
E)(-$\infty$, $\infty$ )

Question

The interval of convergence of the power series $\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { n ^ { 2 } + 3 }$ is

A)(-1,1)
B)[-1,1)
C)(-1,1]
D)[-1,1]
E)(-$\infty$, $\infty$ )

Quizplus · Accepted Answer

The interval of convergence for the given power series can be found using the ratio test, which involves taking the limit as $n$ approaches infinity of the absolute value of the ratio of the $(n+1)$th term to the $n$th term of the series. For the given series, this ratio simplifies to $|x|$, indicating that the series converges when $|x| < 1$, which corresponds to the interval $(-1, 1)$. However, we must also check the endpoints $x = -1$ and $x = 1$ to determine if the series converges at these points. At $x = 1$, the series becomes $\sum_{n=0}^{\infty} \frac{1}{n^2+3}$, which converges by the comparison test (comparing it to the p-series with $p = 2$, which is known to converge). At $x = -1$, the series also converges for the same reason, as the absolute value of the terms is the same as for $x = 1$. Therefore, the interval of convergence is $[-1, 1]$.

The Interval of Convergence of the Power Series ∑n=0∞xnn2+3\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { n ^ { 2 } + 3 }∑n=0∞​n2+3xn​

The Interval of Convergence of the Power Series $\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { n ^ { 2 } + 3 }$