Math Problem Statement

Suppose that βˆ‘βˆžπ‘›=0π‘Žπ‘›π‘₯π‘›βˆ‘n=0∞anxn converges for π‘₯=2x=2 and diverges for π‘₯=βˆ’4.x=βˆ’4. What can be said about the convergence or divergence of the following series?

βˆ‘βˆžπ‘›=0(βˆ’2)π‘›π‘Žπ‘›andβˆ‘βˆžπ‘›=0(βˆ’1)𝑛7π‘›π‘Žπ‘›

Solution

Given that the power series βˆ‘n=0∞anxn\sum_{n=0}^{\infty} a_n x^n converges at x=2x = 2 and diverges at x=βˆ’4x = -4, we can infer the following about the series:

1. Analyzing the Radius of Convergence

The power series βˆ‘n=0∞anxn\sum_{n=0}^{\infty} a_n x^n has a radius of convergence RR. For the series to converge at x=2x = 2 and diverge at x=βˆ’4x = -4, the radius of convergence RR must satisfy: 2≀R<42 \leq R < 4 This is because the series converges when ∣x∣<R|x| < R and diverges when ∣x∣>R|x| > R. Since ∣2∣=2|2| = 2 is within the radius of convergence, the series converges at x=2x = 2. Conversely, since βˆ£βˆ’4∣=4|-4| = 4 is beyond the radius of convergence, the series diverges at x=βˆ’4x = -4.

2. Convergence of βˆ‘n=0∞(βˆ’2)nan\sum_{n=0}^{\infty} (-2)^n a_n

To determine the convergence of the series βˆ‘n=0∞(βˆ’2)nan\sum_{n=0}^{\infty} (-2)^n a_n, we can substitute x=βˆ’2x = -2 into the original series βˆ‘n=0∞anxn\sum_{n=0}^{\infty} a_n x^n. This transforms the series into: βˆ‘n=0∞an(βˆ’2)n\sum_{n=0}^{\infty} a_n (-2)^n This is the same as evaluating the original series at x=βˆ’2x = -2. Since βˆ£βˆ’2∣=2|-2| = 2 is within the radius of convergence RR, the series βˆ‘n=0∞(βˆ’2)nan\sum_{n=0}^{\infty} (-2)^n a_n converges.

3. Convergence of βˆ‘n=0∞(βˆ’1)n7nan\sum_{n=0}^{\infty} (-1)^n 7^n a_n

To determine the convergence of the series βˆ‘n=0∞(βˆ’1)n7nan\sum_{n=0}^{\infty} (-1)^n 7^n a_n, we substitute x=βˆ’7x = -7 into the original series βˆ‘n=0∞anxn\sum_{n=0}^{\infty} a_n x^n. This transforms the series into: βˆ‘n=0∞an(βˆ’7)n\sum_{n=0}^{\infty} a_n (-7)^n This is equivalent to evaluating the original series at x=βˆ’7x = -7. Since βˆ£βˆ’7∣=7|-7| = 7 is greater than the radius of convergence RR, the series βˆ‘n=0∞(βˆ’1)n7nan\sum_{n=0}^{\infty} (-1)^n 7^n a_n diverges.

Summary

  • βˆ‘n=0∞(βˆ’2)nan\sum_{n=0}^{\infty} (-2)^n a_n converges.
  • βˆ‘n=0∞(βˆ’1)n7nan\sum_{n=0}^{\infty} (-1)^n 7^n a_n diverges.

Would you like more details on this analysis or have any questions?

Relative Questions:

  1. How do you determine the radius of convergence for a power series?
  2. What is the difference between absolute and conditional convergence?
  3. How does the root test relate to the radius of convergence?
  4. What is the significance of the radius of convergence in practical applications?
  5. Can a power series converge outside its radius of convergence under any circumstances?

Tip:

When working with power series, always pay attention to the radius of convergence, as it plays a crucial role in determining where the series converges and where it diverges.

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Math Problem Analysis

Mathematical Concepts

Power series
Radius of convergence
Series convergence and divergence

Formulas

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Theorems

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Suitable Grade Level

Advanced undergraduate level