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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.

Find the values of x for which the series converges: ∑ (7(x/6)^n) from n=0 to ∞.

Find out how to determine the values of x for which the geometric series ∑ (7(x/6)^n) from n=0 to ∞ converges. This problem involves geometric series, using the formula for convergence |r| < 1. Suitable for advanced high school or early college-level students.

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