Question

Determine the radius of convergence of the given power series. $$\sum_{n=0}^{\infty}(x-3)^{n}$$

Short answer

Answer: The radius of convergence (R) for the given power series is 1, and it converges when x is in the interval (2, 4).

Step-by-step solution

Write down the series

The given power series is: $$\sum_{n=0}^{\infty}(x-3)^{n}$$

Apply the ratio test

We will use the ratio test which requires us to find the limit of the absolute value of the ratio of consecutive terms as n approaches infinity: $$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$$ For our series, the nth term (a_n) is \((x-3)^n\). So, the n+1th term (a_(n+1)) is \((x-3)^{n+1}\).

Find the limit

Write down the limit expression and simplify: $$\lim_{n\rightarrow\infty}\left|\frac{(x-3)^{n+1}}{(x-3)^n}\right|$$ Divide the exponents: $$\lim_{n\rightarrow\infty}\left|(x-3)\right|$$

Determine the radius of convergence

For the series to converge, the limit must be less than 1. So we have: $$\left|(x-3)\right|<1$$ This implies that the radius of convergence (R) is 1. The given power series converges when x is in the interval \((2,4)\).

Key concepts

Power Series

A power series is an infinite sum of terms that take the form of \(a_n(x-c)^n\), where \(a_n\) represents the coefficient of the nth term, \(x\) is a variable, and \(c\) is a constant that centers the series. Power series can be thought of as polynomials with infinitely many terms, and they play a crucial role in various areas of mathematics, especially in analysis and function approximation.

Each power series has an associated interval, or in some cases a radius, within which it converges to a specific value. Outside of this interval, the series may diverge or not be defined at all. The center \(c\) and the interval of convergence provide crucial information about the behavior of the series and its functional values.

In the exercise, the given power series is centered at \(c=3\) and takes the simplified form \(\sum_{n=0}^{\infty}(x-3)^n\), which is a geometric series with the ratio \(x-3\). To determine where this series converges, we look for the radius of convergence, which tells us the distance from the center \(c\) within which the series converges.

Ratio Test

The Ratio Test is a criterion for determining whether an infinite series converges or diverges by examining the limit of the absolute value of the ratio of consecutive terms. It is expressed as:

\[\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|\]

If the limit is less than one, the series converges absolutely. If it is greater than one, or if the limit does not exist, the series diverges. If the limit equals one, the test is inconclusive, and alternative methods must be employed.

Applying the ratio test to the provided exercise, we aim to find the limit of consecutive terms \(\left|(x-3)^{n+1}/(x-3)^n\right|\), which simplifies to \(\left|x-3\right|\). Here, the test conclusively determines the radius of convergence because the limit is a fixed value and does not depend on \(n\). Thus, by setting the absolute value less than one, we establish the conditions for the series' convergence.

Limit of a Sequence

The limit of a sequence is the value that the sequence's terms approach as the index \(n\) increases indefinitely. Mathematically, we denote the limit of a sequence \(\{a_n\}\) as\(n\) approaches infinity by:

\[\lim_{n\rightarrow\infty}a_n\]

If the limit exists and is finite, the sequence is said to converge; if not, it diverges. This concept is pivotal in calculus and analysis as it forms the foundation for the convergence or divergence of sequences and series.

In the context of the exercise, finding the limit of the sequence of consecutive terms' ratios \(\left|(x-3)^{n+1}/(x-3)^n\right|\) doesn't depend on \(n\), since it simplifies to \(\left|x-3\right|\) and is thus independent of the sequence index. It indicates that the behavior of the power series is determined by the value of \(x\) in relation to its distance from the center, \(c=3\). Therefore, the limit here is used to define the condition for convergence of the power series, namely, \(\left|x-3\right|<1\), which leads us to determine the radius of convergence as 1.