Math Problem Statement

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

Solution

To analyze the geometric series n=1(x3)n5n\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}, we follow these steps:

Part (a): Determine all values of xx for which the series is convergent

  1. Identify the series: n=1(x3)n5n\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}

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

  3. Rewrite the given series in geometric series form: Here, the first term aa is 0 (since the series starts from n=1n=1), and the common ratio rr is: r=x35r = \frac{x-3}{5}

  4. Determine the condition for convergence: For the series to converge, the common ratio must satisfy: x35<1\left|\frac{x-3}{5}\right| < 1

  5. Solve the inequality: x35<1    1<x35<1\left|\frac{x-3}{5}\right| < 1 \implies -1 < \frac{x-3}{5} < 1 Multiply through by 5: 5<x3<5-5 < x-3 < 5 Add 3 to all parts of the inequality: 2<x<8-2 < x < 8

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

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

  1. Sum of a convergent geometric series: The sum SS of an infinite geometric series n=0arn\sum_{n=0}^{\infty} ar^n with r<1|r| < 1 is given by: S=a1rS = \frac{a}{1-r}

  2. Apply the formula to the given series: For our series n=1(x3)n5n\sum_{n=1}^{\infty} \frac{(x-3)^n}{5^n}, we recognize that the series starts at n=1n=1, not n=0n=0. Therefore, we need to adjust our sum formula accordingly: S=n=1(x35)n=(x35)1(x35)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=x351x35=x355(x3)5=x35(x3)=x38x+3=x311xS = \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<8-2 < x < 8, the sum of the series is x311x\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=0n = 0 instead of n=1n = 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|r| < 1 for convergence and use the sum formula to find the exact sum for the given interval.

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Math Problem Analysis

Mathematical Concepts

Geometric Series
Convergence Criteria

Formulas

Sum of a Geometric Series

Theorems

Geometric Series Convergence Criteria

Suitable Grade Level

Grades 11-12