Problem 29 - Essential Calculus

Question

In Exercises 29 and \(30,\) find the radius of convergence of the power series, where \(c>0\) and \(k\) is a positive integer. $$ \sum_{n=1}^{\infty} \frac{(x-c)^{n-1}}{c^{n-1}} $$

Step-by-Step Solution

Answer

The radius of convergence of the power series is \(c\).

Step 1: Apply the ratio test

For applying the ratio test, calculate the ratio of the \((n+1)^{th}\) term to the \(n^{th}\) term of the series. In this case, it looks like this: \[ \left|\frac{\frac{(x-c)^n}{c^n}}{\frac{(x-c)^{n-1}}{c^{n-1}}}\right| = \left|\frac{(x-c)}{c}\right| \]

Step 2: Evaluate the limit

Next, find the limit of the absolute value of the ratio as \(n\) approaches infinity: \[ \lim_{{n \to \infty}} \left|\frac{(x-c)}{c}\right| = \left|\frac{x-c}{c}\right| \]Therefore, \((x-c)/c\) is independent of \(n\), the limit does not depend on \(n\).

Step 3: Determine the radius of convergence

The series converges when the limit from the ratio test is less than 1:\[ \left|\frac{x-c}{c}\right| < 1 \]This can be rewritten as \(-1 < (x-c)/c < 1\), or \(-c < x-c < c\), and finally \(0 < x < 2c\). Therefore, the radius of convergence is \(c\), which is half the interval of convergence.

Key Concepts

Power Series

In mathematics, a power series is a series of the form:

• \[ \sum_{n=0}^{\infty} a_n(x-c)^n \]

Here, \(a_n\) represents the coefficients of the series, \(x\) is the variable of the series, and \(c\) is the center of the series. In simpler terms, a power series is an infinite polynomial. The series can converge or diverge, depending on the value of \(x\). Power series are particularly useful because they allow us to approximate functions that might otherwise be difficult to manage, especially in calculus and differential equations.
This allows mathematicians to work with functions in a different, sometimes easier form. Understanding where a power series converges is crucial in determining its usefulness in practical applications.

Ratio Test

The ratio test is a method for testing the convergence of infinite series. Specifically, it is used to determine if a series converges absolutely. For a series \(\sum_{n=1}^{\infty} a_n\), the ratio test considers the limit:

• \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

The series converges absolutely if \(L < 1\) and diverges if \(L > 1\).
If \(L = 1\), the test is inconclusive. The key idea behind the ratio test is to compare the terms of the series to a geometric series, where a similar ratio determines convergence.
This makes the ratio test a powerful tool for analyzing where series converge, particularly when dealing with power series.

Interval of Convergence

The interval of convergence is the set of \(x\) values for which a power series converges. To find this interval, we first use the radius of convergence determined by the ratio test. Once the radius is known, the interval can be expressed as follows:

• \( \text{Interval: } (c-R, c+R) \),

where \(c\) is the center of the series and \(R\) is the radius of convergence.
Additionally, both endpoints of the interval need to be checked individually since the series may converge or diverge at these points. Understanding the interval of convergence helps us to know where the series can be effectively used to represent a function.

Limit of a Sequence

The limit of a sequence is the value that the terms of a sequence "approach" as the number of terms goes to infinity. In mathematical terms, a sequence \(a_n\) has a limit \(L\) if, for every positive number \(\epsilon\), there exists a natural number \(N\) such that for all \(n > N\), the terms of the sequence satisfy:

• \[ |a_n - L| < \epsilon \]

This means that as \(n\) becomes very large, \(a_n\) gets arbitrarily close to \(L\). Understanding the concept of limits is crucial not only in calculus but also in analyzing the convergence properties of series and sequences. The ratio test, for example, heavily relies on finding the limit of the ratio of consecutive terms to determine convergence behavior.