Math Problem Statement

Solve it step-by-step solutions (Power Series)

Solve it step-by-step solutions (Power Series)

Solution

Problem 1:

The series is:

n=1(x5)n10n33.\sum_{n=1}^\infty \frac{(x-5)^n}{10^{n^3 - 3}}.

Step-by-Step Solution:

  1. General Term Identification: The general term of the series is: an=(x5)n10n33.a_n = \frac{(x-5)^n}{10^{n^3 - 3}}.

  2. Convergence Test: To determine convergence, we apply the ratio test: limnan+1an<1.\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1. Substituting ana_n, we get: an+1an=(x5)n+110(n+1)3310n33(x5)n.\frac{a_{n+1}}{a_n} = \frac{(x-5)^{n+1}}{10^{(n+1)^3 - 3}} \cdot \frac{10^{n^3 - 3}}{(x-5)^n}. Simplify: an+1an=x510n3310(n+1)33.\frac{a_{n+1}}{a_n} = |x-5| \cdot \frac{10^{n^3 - 3}}{10^{(n+1)^3 - 3}}. The exponential term simplifies as: 10n3(n+1)3=103n23n1.10^{n^3 - (n+1)^3} = 10^{-3n^2 - 3n - 1}. Therefore: an+1an=x5103n23n1.\frac{a_{n+1}}{a_n} = |x-5| \cdot 10^{-3n^2 - 3n - 1}.

  3. Limit: As nn \to \infty, the term 103n23n110^{-3n^2 - 3n - 1} approaches 00, regardless of xx. Hence, the series always converges.


Problem 2:

The series is:

n=1(x2)n3nn2.\sum_{n=1}^\infty \frac{(x-2)^n}{3^n \cdot n^2}.

Step-by-Step Solution:

  1. General Term Identification: The general term of the series is: an=(x2)n3nn2.a_n = \frac{(x-2)^n}{3^n \cdot n^2}.

  2. Convergence Test: Use the ratio test: limnan+1an<1.\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1. Substituting ana_n, we get: an+1an=(x2)n+13n+1(n+1)23nn2(x2)n.\frac{a_{n+1}}{a_n} = \frac{(x-2)^{n+1}}{3^{n+1} \cdot (n+1)^2} \cdot \frac{3^n \cdot n^2}{(x-2)^n}. Simplify: an+1an=x23n2(n+1)2.\frac{a_{n+1}}{a_n} = \frac{|x-2|}{3} \cdot \frac{n^2}{(n+1)^2}. As nn \to \infty, the factor n2(n+1)21\frac{n^2}{(n+1)^2} \to 1. Thus: limnan+1an=x23.\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{|x-2|}{3}.

  3. Radius of Convergence: For convergence: x23<1.\frac{|x-2|}{3} < 1. Solve for xx: x2<3.|x-2| < 3. This gives the interval of convergence: 1<x<5.-1 < x < 5.

  4. Boundary Check: At x=1x = -1 and x=5x = 5, the series becomes: n=11n2,\sum_{n=1}^\infty \frac{1}{n^2}, which converges (p-series, p=2>1p = 2 > 1). Hence, the series converges at both boundaries.

  5. Final Answer: The interval of convergence is: [1,5].[-1, 5].


Would you like further clarification on any of these steps? Here are 5 related questions:

  1. How does the ratio test determine convergence of a series?
  2. Can the radius of convergence change for different power series?
  3. What is the significance of boundary checks in convergence problems?
  4. How do you simplify exponential terms in convergence tests?
  5. Why does the p-series test apply to these boundaries?

Tip: Always simplify terms as much as possible before applying convergence tests for clarity.

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Math Problem Analysis

Mathematical Concepts

Power Series
Ratio Test
Convergence of Infinite Series

Formulas

General Term of a Series: a_n
Ratio Test: lim(n→∞) |a_(n+1) / a_n| < 1
P-Series Convergence: sum(1/n^p) converges if p > 1

Theorems

Ratio Test
P-Series Test

Suitable Grade Level

Grades 11-12 or College Level