To solve the problem of finding the values of $x$ so that the given infinite series converges, we analyze the series:

$\sum_{n=0}^\infty \frac{(x+1)^n}{13^n}.$

This is a geometric series with the general term: $a_n = \frac{(x+1)^n}{13^n}.$

For a geometric series to converge, the absolute value of the common ratio $r$ must satisfy: $|r| < 1.$

Here, the common ratio $r$ is: $r = \frac{x+1}{13}.$

For convergence: $\left|\frac{x+1}{13}\right| < 1.$

Step 1: Simplify the inequality

$-1 < \frac{x+1}{13} < 1.$

Multiply through by 13: $-13 < x+1 < 13.$

Step 2: Solve for $x$

Subtract 1 from all parts of the inequality: $-14 < x < 12.$

Final Answer:

The series converges for $x \in (-14, 12)$.

Would you like me to explain any specific step or expand on how this relates to other types of series convergence? Here are 5 related questions:

1. What happens if the series were not geometric?
2. How does the ratio test help confirm convergence here?
3. Can the convergence interval be expanded if the series changes?
4. What does the convergence behavior look like at $x = -14$ and $x = 12$?
5. How does this apply to power series and their radius of convergence?

Tip: Always check endpoints of an interval to see if they are included when dealing with series convergence!