Consider the geometric series sigma n=1 to infinity (x-3)^n/5^n (a) determine all values of x for which series is convergent, justify answer. (b) For the values of x which you found in part(a) what is the sum of series

To analyze the geometric series \(\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}\), we follow these steps:

### Part (a): Determine all values of \( x \) for which the series is convergent

1. **Identify the series:**
   \[
   \sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}
   \]

2. **Recognize the form of a geometric series:**
   A geometric series \(\sum_{n=0}^{\infty} ar^n\) converges if and only if \(|r| < 1\).

3. **Rewrite the given series in geometric series form:**
   Here, the first term \(a\) is 0 (since the series starts from \(n=1\)), and the common ratio \(r\) is:
   \[
   r = \frac{x-3}{5}
   \]

4. **Determine the condition for convergence:**
   For the series to converge, the common ratio must satisfy:
   \[
   \left|\frac{x-3}{5}\right| < 1
   \]

5. **Solve the inequality:**
   \[
   \left|\frac{x-3}{5}\right| < 1 \implies -1 < \frac{x-3}{5} < 1
   \]
   Multiply through by 5:
   \[
   -5 < x-3 < 5
   \]
   Add 3 to all parts of the inequality:
   \[
   -2 < x < 8
   \]

Thus, the series converges for \(\boxed{-2 < x < 8}\).

### Part (b): Find the sum of the series for values of \( x \) in the interval \(-2 < x < 8\)

1. **Sum of a convergent geometric series:**
   The sum \(S\) of an infinite geometric series \(\sum_{n=0}^{\infty} ar^n\) with \(|r| < 1\) is given by:
   \[
   S = \frac{a}{1-r}
   \]

2. **Apply the formula to the given series:**
   For our series \(\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}\), we recognize that the series starts at \(n=1\), not \(n=0\). Therefore, we need to adjust our sum formula accordingly:
   \[
   S = \sum_{n=1}^{\infty} \left(\frac{x-3}{5}\right)^n = \frac{\left(\frac{x-3}{5}\right)}{1 - \left(\frac{x-3}{5}\right)}
   \]

3. **Simplify the expression:**
   \[
   S = \frac{\frac{x-3}{5}}{1 - \frac{x-3}{5}} = \frac{\frac{x-3}{5}}{\frac{5-(x-3)}{5}} = \frac{x-3}{5 - (x-3)} = \frac{x-3}{8-x+3} = \frac{x-3}{11-x}
   \]

Thus, for \(-2 < x < 8\), the sum of the series is \(\boxed{\frac{x-3}{11-x}}\).

Would you like more detailed explanations or have any specific questions?

Here are some related questions to expand your understanding:

1. What is the geometric series test for convergence?
2. How do you find the sum of a finite geometric series?
3. How does the radius of convergence relate to the interval of convergence for a power series?
4. What are the differences between geometric and arithmetic series?
5. Can a geometric series converge to zero?
6. What is the significance of the common ratio in determining convergence?
7. How do you determine the sum of a geometric series if the series starts at \( n = 0 \) instead of \( n = 1 \)?
8. What are some applications of geometric series in real-world problems?

**Tip:** When working with geometric series, always ensure the common ratio \( |r| < 1 \) for convergence and use the sum formula to find the exact sum for the given interval.

Explore the convergence criteria and sum calculation of the geometric series \(\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}\). Learn how to determine the values of x for convergence and find the sum of the series within the interval \(-2 < x < 8\).

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