The given series is:

$\sum_{n=0}^{\infty} 7 \left(\frac{x}{6}\right)^n$

This is a geometric series of the form:

$\sum_{n=0}^{\infty} ar^n$

where $a = 7$ and $r = \frac{x}{6}$. A geometric series converges if and only if the absolute value of the common ratio $r$ is less than 1, that is:

$|r| < 1$

In this case:

$\left|\frac{x}{6}\right| < 1$

To solve for $x$:

$-1 < \frac{x}{6} < 1$

Multiplying through by 6:

$-6 < x < 6$

Thus, the series converges when $x$ is in the interval:

$(-6, 6)$

Final Answer:

The series converges for $x \in (-6, 6)$.

Would you like further details or have any questions?

Related Questions:

1. How would the answer change if the series started at $n = 1$ instead of $n = 0$?
2. What happens to the series for $x = -6$ or $x = 6$?
3. Can you explain the sum of a geometric series?
4. How do you apply the ratio test to determine the convergence of a series?
5. How would the radius of convergence change if the denominator were different?

Tip: When dealing with geometric series, always check the condition $|r| < 1$ for convergence.