Math Problem Statement

Determine interval of convergence for the series: ∑ (3^k(x - 4)^k) / k.

Solution

We are given the series

k=03k(x4)kk\sum_{k=0}^{\infty} \frac{3^k (x - 4)^k}{k}

and asked to find its interval of convergence.

Step 1: Apply the Ratio Test

To find the interval of convergence, we use the Ratio Test for the general term ak=3k(x4)kka_k = \frac{3^k (x - 4)^k}{k}. According to the ratio test, for a series ak\sum a_k, the radius of convergence RR can be determined by checking the limit:

L=limkak+1ak.L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|.

We will now compute LL for the given series.

Step 2: Compute the Ratio ak+1ak\left| \frac{a_{k+1}}{a_k} \right|

First, write the terms ak+1a_{k+1} and aka_k:

  • ak=3k(x4)kka_k = \frac{3^k (x - 4)^k}{k},
  • ak+1=3k+1(x4)k+1k+1=33k(x4)(x4)kk+1a_{k+1} = \frac{3^{k+1} (x - 4)^{k+1}}{k+1} = \frac{3 \cdot 3^k (x - 4) \cdot (x - 4)^k}{k+1}.

Now, take the ratio:

ak+1ak=33k(x4)(x4)kk+13k(x4)kk=3(x4)1kk+1.\frac{a_{k+1}}{a_k} = \frac{\frac{3 \cdot 3^k (x - 4) \cdot (x - 4)^k}{k+1}}{\frac{3^k (x - 4)^k}{k}} = \frac{3(x - 4)}{1} \cdot \frac{k}{k+1}.

Simplifying this:

ak+1ak=3x4kk+1.\left| \frac{a_{k+1}}{a_k} \right| = 3 |x - 4| \cdot \frac{k}{k+1}.

As kk \to \infty, kk+11\frac{k}{k+1} \to 1, so we have:

L=limkak+1ak=3x4.L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = 3 |x - 4|.

Step 3: Apply the Ratio Test Condition

The Ratio Test states that the series converges when L<1L < 1. Thus, we require:

3x4<1.3 |x - 4| < 1.

Dividing both sides by 3:

x4<13.|x - 4| < \frac{1}{3}.

Step 4: Solve for the Interval of Convergence

This inequality means that xx must lie within 13\frac{1}{3} units of 4. Therefore, the interval is:

413<x<4+13,4 - \frac{1}{3} < x < 4 + \frac{1}{3},

or equivalently:

113<x<133.\frac{11}{3} < x < \frac{13}{3}.

Step 5: Check the Endpoints

To determine if the endpoints x=113x = \frac{11}{3} and x=133x = \frac{13}{3} are included in the interval, we substitute these values into the series and check for convergence.

  • At x=113x = \frac{11}{3}, the series becomes a harmonic series, which diverges.
  • At x=133x = \frac{13}{3}, the series again becomes harmonic, and it diverges.

Thus, the endpoints are not included in the interval.

Final Answer:

The interval of convergence is:

(113,133).\left( \frac{11}{3}, \frac{13}{3} \right).

Would you like a more detailed explanation of any part, or have any further questions?

Here are 5 related questions to explore this topic further:

  1. How do you check convergence at endpoints using other convergence tests?
  2. What happens if the Ratio Test gives a limit equal to 1?
  3. How does the Ratio Test work for other types of series, like alternating series?
  4. What is the difference between absolute and conditional convergence?
  5. How can power series be applied in approximating functions?

Tip: For any power series, always test both endpoints to ensure whether they are included in the interval of convergence!

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Math Problem Analysis

Mathematical Concepts

Power Series
Convergence Tests
Ratio Test

Formulas

Ratio Test: L = lim (k → ∞) |a_(k+1) / a_k|
|x - 4| < 1/3

Theorems

Ratio Test for Series Convergence

Suitable Grade Level

College-Level Calculus