Find the interval of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } ( - 1 ) ^ { k } \frac { k ^ { 2 } } { 2 ^ { k } } ( x - 2 ) ^ { k }$$
A) [0, 4)
B) (0, 4)
C) (0, 4]
D) [0, 4]

Question

Quizplus · Accepted Answer

The series is an alternating series, so we can use the Alternating Series Test. The terms of the series are decreasing because $k^2/2^k$ is decreasing for $k\ge 1$. So the series converges by the Alternating Series Test.To find the interval of convergence, we use the Ratio Test. The limit of the ratio of consecutive terms is$$\lim_{k	o\infty} \left| \frac{a_{k+1}}{a_k} ight| = \lim_{k	o\infty} \left| \frac{(-1)^{k+1} (k+1)^2/2^{k+1} (x-2)^{k+1}}{(-1)^k k^2/2^k (x-2)^k} ight| = \lim_{k	o\infty} \left| \frac{(k+1)^2}{k^2} \cdot \frac{2^k}{2^{k+1}} \cdot \frac{1}{x-2} ight| = \lim_{k	o\infty} \left| \frac{k^2+2k+1}{k^2} \cdot \frac{1}{2} \cdot \frac{1}{x-2} ight| = \frac{1}{2|x-2|}$$So the series converges if $\frac{1}{2|x-2|}<1$, which is equivalent to $|x-2|>2$. Therefore, the interval of convergence is $(0,4)$.

Find the Interval of Convergence of the Power Series ∑k=0∞(−1)kk22k(x−2)k\sum _ { k = 0 } ^ { \infty } ( - 1 ) ^ { k } \frac { k ^ { 2 } } { 2 ^ { k } } ( x - 2 ) ^ { k }k=0∑∞​(−1)k2kk2​(x−2)k

Find the Interval of Convergence of the Power Series $\sum _ { k = 0 } ^ { \infty } ( - 1 ) ^ { k } \frac { k ^ { 2 } } { 2 ^ { k } } ( x - 2 ) ^ { k }$