For what values of p does the series $1+9 p+(9 p)^{2}+(9 p)^{3}+\ldots+(9 p)^{n}+$ converge, if any?
A) $\frac{-1}{9}$$

Question

For what values of p does the series $1+9 p+(9 p)^{2}+(9 p)^{3}+\ldots+(9 p)^{n}+$ converge, if any?
A) $\frac{-1}{9}$$<$$p<$$\frac{1}{9}$
B) $\frac{-1}{18}$ $<$$p<$$\frac{1}{18}$
C)-1 < p < 1
D) $-1 \leq p \leq 1$
E)The series diverges for all values of p.

Quizplus · Accepted Answer

The series given is a geometric series with the first term $a = 1$ and common ratio $r = 9p$. For a geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., $|9p| < 1$. Solving this inequality gives $\frac{-1}{9} < p < \frac{1}{9}$.

For What Values of P Does the Series 1+9p+(9p)2+(9p)3+…+(9p)n+1+9 p+(9 p)^{2}+(9 p)^{3}+\ldots+(9 p)^{n}+1+9p+(9p)2+(9p)3+…+(9p)n+

For What Values of P Does the Series $1+9 p+(9 p)^{2}+(9 p)^{3}+\ldots+(9 p)^{n}+$