Chapter 11: Problem 20
Find the interval of convergence of the power series. $$ \sum_{n=1}^{\infty} \frac{1}{n 5^{n}}(x-5)^{n} $$
Short Answer
Expert verified
The interval of convergence is \([0, 10)\).
Step by step solution
01
Identify the General Form of the Power Series
The given power series is \( \sum_{n=1}^{\infty} \frac{1}{n 5^{n}}(x-5)^{n} \). We identify the general term of this series as \( a_n = \frac{1}{n 5^{n}}(x-5)^{n} \).
02
Apply the Ratio Test for Convergence
To find the interval of convergence, we use the Ratio Test. We compute the absolute value of the ratio of consecutive terms:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{1}{(n+1) 5^{n+1}} \cdot \frac{n 5^{n}}{1} \cdot (x-5) \right|\]Simplifying, we have:\[\left| \frac{x-5}{5} \right| \cdot \frac{n}{n+1} \]
03
Evaluate the Limit of the Ratio Test
Consider the limit as \( n \to \infty \):\[\lim_{n \to \infty} \left| \frac{x-5}{5} \right| \cdot \frac{n}{n+1} = \left| \frac{x-5}{5} \right| \cdot \lim_{n \to \infty} \frac{n}{n+1} = \left| \frac{x-5}{5} \right| \cdot 1\]Thus, the limit is \( \left| \frac{x-5}{5} \right| \).
04
Determine the Interval of Convergence
The Ratio Test states that the series converges if the limit is less than 1. Thus, solve the inequality:\[\left| \frac{x-5}{5} \right| < 1\]This results in:\[-1 < \frac{x-5}{5} < 1\]Multiply through by 5:\[-5 < x-5 < 5\]Adding 5 to all sides:\[0 < x < 10\]
05
Check the Endpoints for Convergence
Evaluate the series at the endpoints, \( x = 0 \) and \( x = 10 \).- At \( x = 0 \): \[ \sum_{n=1}^{\infty} \frac{1}{n \, 5^{n}} (-5)^{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \] Is the alternating harmonic series, which converges.- At \( x = 10 \): \[ \sum_{n=1}^{\infty} \frac{1}{n \, 5^{n}} (5)^{n} = \sum_{n=1}^{\infty} \frac{1}{n} \] This is the harmonic series, which diverges.
06
Summarize the Interval of Convergence
Since the series converges at \( x = 0 \) but diverges at \( x = 10 \), the interval of convergence is \([0, 10)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \). This means you have a sum of terms, each involving a constant \( a_n \), and a variable expression \( (x-c)^n \). Here, \( c \) is called the center of the series and \( x \) is a variable. The power series can converge or diverge, depending on the value of \( x \). Determining where it converges (called the interval of convergence) is a common task. In the exercise, our power series is centered at 5, which is reflected in the expression \( (x-5)^n \), and each term is combined with a factor of \( \frac{1}{n 5^n} \). Understanding the general form is key to analyzing convergence and finding the interval.
Ratio Test
The ratio test is a popular method for checking the convergence of a series. For a series \( \sum a_n \), we examine \( \left| \frac{a_{n+1}}{a_n} \right| \). If the limit of this ratio as \( n \to \infty \) is less than 1, the series converges absolutely. If the limit is greater than 1 or infinite, the series diverges. If the limit equals 1, the test is inconclusive, and other tests may be needed. For the exercise, we calculated the ratio between consecutive terms using \( a_n = \frac{1}{n 5^n}(x-5)^n \). It gave us \( \left| \frac{x-5}{5} \right| \), which we set less than 1 to determine converge.
Convergence
Convergence in terms of series happens when adding up the infinitely many terms of a series results in a finite number. Mathematically, for a series \( \sum a_n \), it converges if the sequence of partial sums \( S_n \) approaches a finite limit as \( n \to \infty \). The interval of convergence specifies the values of \( x \) for which the power series converges. In our scenario, we solved the inequality \( \left| \frac{x-5}{5} \right| < 1 \) which led us to the interval \( 0 < x < 10 \). The endpoints \( x = 0 \) and \( x = 10 \) require individual checks to determine convergence there as well.
Alternating Harmonic Series
The alternating harmonic series is given by \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \). This series is special because it converges even though the absolute value of its terms form a divergent series (the regular harmonic series, \( \sum \frac{1}{n} \)). The convergence of the alternating harmonic series is confirmed using the Alternating Series Test which states that for a series \( \sum (-1)^n b_n \), if \( b_n \) is a sequence of positive terms that decrease to zero, the series converges. In the exercise, when \( x = 0 \), the power series becomes an alternating harmonic series, which helps us declare convergence at that end point.