For what values of x does the series converge conditionally?
-$$\sum _ { n = 1 } ^ { \infty } \frac { ( x + 7 ) ^ { n } } { \sqrt { n } }$$
A) x = -6
B) x = -8, x = -6
C) x = -8
D) None

Question

Quizplus · Accepted Answer

To determine the values of $x$ for which the series converges conditionally, we use the alternating series test. Thus, we need to check two conditions:

1. The terms of the series $\left(\frac{(x+7)^n}{\sqrt{n}}ight)$ are decreasing in absolute value for all $n\ge 1$.
2. $\lim_{n	o\infty}\frac{(x+7)^n}{\sqrt{n}}=0$.

For the first condition, we have:

$$\left|\frac{(x+7)^{n+1}}{\sqrt{n+1}}ight|<\left|\frac{(x+7)^n}{\sqrt{n}}ight|\iff \left|\frac{x+7}{\sqrt{n+1}}ight|<1$$

which is satisfied for all $x\le -8$.

For the second condition, we have:

$$\lim_{n	o\infty}\frac{(x+7)^n}{\sqrt{n}}=\lim_{n	o\infty}(x+7)^n\cdot \frac{1}{\sqrt{n}}=\begin{cases} 0 & 	ext{if } x<-7\+\infty & 	ext{if } x>-7\end{cases}$$

Thus, the only $x$ value that satisfies both conditions is $x=-8$, so the series converges conditionally for $x=-8$.

For What Values of X Does the Series Converge Conditionally ∑n=1∞(x+7)nn\sum _ { n = 1 } ^ { \infty } \frac { ( x + 7 ) ^ { n } } { \sqrt { n } }n=1∑∞​n​(x+7)n​

For What Values of X Does the Series Converge Conditionally $\sum _ { n = 1 } ^ { \infty } \frac { ( x + 7 ) ^ { n } } { \sqrt { n } }$