Find the values of $x$ for which the series converges. (Enter your answer using interval notation.) $\displaystyle\sum_{n=1}^\infty (x + 6)^n.$

$\boxed{?}$

Find the sum of the series for those values of $x$.

$\boxed{?}$

Neetesh Kumar | December 28, 2024

Calculus Homework Help

This is the solution to Math 1c
Assignment: 11.2 Question Number 22
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Step-by-step solution:

Step 1: Analyze the series:

The given series is:

$\displaystyle\sum_{n=1}^\infty (x+6)^n.$

This is a geometric series with:

• First term ($a$): $(x+6)$ (when $n=1$),
• Common ratio ($r$): $(x+6)$.

A geometric series converges if and only if the absolute value of the common ratio $r$ satisfies:

$|r| < 1.$

Substitute $r = x+6$:

$|x+6| < 1.$

Simplify the inequality:

$-1 < x+6 < 1.$

Solve for $x$:

$-7 < x < -5.$

Thus, the series converges for:

$x \in (-7, -5).$

Step 2: Find the sum of the series:

When $x \in (-7, -5)$, the series converges, and the sum of a geometric series starting from $n=1$ is:

$S = \frac{r}{1 - r}.$

Here:

$r = x+6.$

Substitute:

$S = \frac{x+6}{1 - (x+6)}.$

Simplify the denominator:

$S = \frac{x+6}{1 - x - 6} = \frac{x+6}{-x - 5}.$

Rewriting with a negative numerator:

$S = \frac{-x - 6}{x + 5}.$

Final Answer:

1. The series converges for:

$x \in \boxed{(-7, -5)}$

2. The sum of the series for these values of $x$ is:

$S = \boxed{\frac{-x - 6}{x + 5}}$

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