Question

Find the function represented by the following series, and find the interval of convergence of the series. (Not all these series are power series.) $$\sum_{k=1}^{\infty} \frac{(x-2)^{k}}{3^{2 k}}$$

Short answer

Answer: The interval of convergence for the series is \((-7, 11)\), excluding the endpoints.

Step-by-step solution

Identifying the General Term of the Series

The given series is: $$\sum_{k=1}^{\infty} \frac{(x-2)^{k}}{3^{2 k}}$$ Here, the general term of the series can be expressed as: $$a_k = \frac{(x-2)^{k}}{3^{2 k}}$$

Apply the Ratio Test to find the Interval of Convergence

The Ratio Test states that for a given series \(\sum_{k=1}^{\infty} a_k\), the series converges if, $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = L < 1$$ Now, let's find the ratio: $$\frac{a_{k+1}}{a_k} = \frac{\frac{(x-2)^{k+1}}{3^{2(k+1)}}}{\frac{(x-2)^{k}}{3^{2 k}}} = \frac{(x-2)^{k+1} \cdot 3^{2 k}}{(x-2)^{k} \cdot 3^{2(k+1)}} = \frac{(x-2)}{3^2}$$ Now, let's find the limit as k approaches infinity: $$L = \lim_{k \to \infty} \left|\frac{(x-2)}{3^2}\right| = \frac{|x-2|}{9}$$ For the series to converge, we must have \(L<1\): $$\frac{|x-2|}{9} < 1$$ Multiplying both sides by 9: $$|x-2|<9$$

Determine Convergence at the Interval Endpoints

From the inequality, we can see that the interval of convergence is \((-7, 11)\). Now, we will test the endpoints to check for convergence. For \(x=-7\), $$\sum_{k=1}^{\infty} \frac{(-9)^{k}}{3^{2 k}}$$ Here, each term alternates in sign and does not tend to zero as k increases, so the series diverges. For \(x=11\), $$\sum_{k=1}^{\infty} \frac{(9)^{k}}{3^{2 k}}$$ This series has terms that tend to zero, but the sum does not converge due to the lack of cancellation of terms as they get larger. Thus, the series diverges.

Final Answer: Interval of Convergence

The interval of convergence for the given series is \((-7, 11)\), excluding the endpoints.

Key concepts

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of an infinite series. It helps you decide if a series converges by examining the ratio between successive terms. If the limit of this ratio, as the term number approaches infinity, is less than one, the series converges.
The Ratio Test is particularly useful for series where the terms involve factorials or other fast-growing expressions.

The process involves the following steps:

• Identify the general term of the series, denoted as \(a_k\).
• Compute the ratio \(\frac{a_{k+1}}{a_k}\).
• Find the limit \(L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).
• Apply the test: if \(L < 1\), the series converges absolutely; if \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In our exercise, the Ratio Test was applied to the series \( \sum_{k=1}^{\infty} \frac{(x-2)^{k}}{3^{2 k}} \). This estimated the interval of values for which the series converges by simplifying the limit to \( \frac{|x-2|}{9} < 1 \), leading us to consider the interval for x.

Interval of Convergence

The interval of convergence describes the range of \(x\)-values for which a given power series converges. It's a key concept when working with series, as it tells you where the series behaves as expected.
To find the interval of convergence, you typically:

• Apply a convergence test, such as the Ratio Test, to establish a basic interval.
• Check the series at the endpoints of this interval, as the test might not be conclusive there.

In the given exercise, after finding that \(\frac{|x-2|}{9} < 1\), we determined the interval \((-7, 11)\). This comes from solving \(|x-2| < 9\), leading to \(-7 < x < 11\). Testing the endpoints, \(x = -7\) and \(x = 11\), revealed that the series diverges at both points, meaning the interval is open exclude its endpoints. Therefore, the series only converges for \(-7 < x < 11\), but not at \(x = -7\) or \(x = 11\).

Power Series

A power series is an infinite series of the form \(\sum_{k=0}^{\infty} a_k(x-c)^k\), where \(a_k\) are coefficients and \(c\) is the center of the series. Power series are like polynomials that extend to infinity, allowing us to represent functions more flexibly.
Each term in a power series is a power of \(x - c\), making it highly adaptable for representing functions around a point \(c\). These series can represent a variety of functions, provided that \(x\) lies within the interval of convergence.

From the exercise, our given series \( \sum_{k=1}^{\infty} \frac{(x-2)^{k}}{3^{2 k}} \) is indeed a power series centered at \(c = 2\). This tells us \(x-2\) is being repeatedly raised to increasing powers, adjusted by a denominator factor \(3^{2k}\), forming each term in the series. Understanding power series and their intervals of convergence allows you to model functions in calculus and many applications of mathematics!