Find the interval of convergence of the power series: $$\sum _ { k = 1 } ^ { \infty } \frac { 1 } { k } ( x + 3 ) ^ { k }$$
A) (-2, 2)
B) (-4, -2)
C) [-4, 2]
D) [-4, -2)

Question

Quizplus · Accepted Answer

The interval of convergence can be found using the ratio test, which involves taking the limit as $k$ approaches infinity of the absolute value of the ratio of the $(k+1)$th term to the $k$th term of the series. For the given series, the ratio is $\left| \frac{x+3}{k+1} \cdot \frac{k}{x+3} \right| = \left| \frac{k}{k+1} \right| \cdot |x+3|$. As $k$ approaches infinity, $\left| \frac{k}{k+1} \right|$ approaches 1, so the ratio test simplifies to $|x+3| < 1$. This inequality implies that $-1 < x + 3 < 1$, or $-4 < x < -2$, which corresponds to the open interval $(-4, -2)$. Therefore, the interval of convergence is $(-4, -2)$, which matches option D.

Find the Interval of Convergence of the Power Series ∑k=1∞1k(x+3)k\sum _ { k = 1 } ^ { \infty } \frac { 1 } { k } ( x + 3 ) ^ { k }k=1∑∞​k1​(x+3)k

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