Problem 4
Question
Find the interval of convergence. $$\sum \frac{1}{k} x^{k}$$
Step-by-Step Solution
Verified Answer
The interval of convergence for the power series \(\sum\frac{1}{k}x^{k}\) is \((-1, 1)\).
1Step 1: Identify the series and key components
Given the power series: \(\sum\frac{1}{k}x^{k}\)
The general term is: \(a_k = \frac{1}{k}x^k\)
2Step 2: Apply the Ratio Test
The Ratio Test is defined as: \(L = \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\)
Now, we will find the ratio of consecutive terms for our power series.
\(\frac{a_{k+1}}{a_{k}} = \frac{\tfrac{1}{k+1} x^{k+1}}{\tfrac{1}{k} x^k}\)
Simplify the expression:
\(\frac{a_{k+1}}{a_{k}} = \frac{k}{k+1}x\)
3Step 3: Find the limit of the ratio
Next, we need to find the limit as k tends to infinity.
\(L = \lim_{k \to \infty} \left|\frac{k}{k+1}x\right|\)
Since the limit of \(\frac{k}{k+1}\) as \(k\) goes to infinity is 1, we are left with:
\(L = |x|\)
4Step 4: Determine interval of convergence
For convergence, the Ratio Test states that \(L < 1\). Therefore, we have:
\(|x| < 1\)
This implies: \(-1 < x < 1\)
So, the interval of convergence for the given power series is (-1, 1).
Key Concepts
Power SeriesRatio TestLimit of a Sequence
Power Series
When we talk about power series, we are referring to an infinite series of the form \[\sum a_k x^k\] where \(a_k\) represents the coefficient of each term, and \(x\) is a variable. The series is centered around a point—often, this point is zero, leading to a series that looks like \(\sum a_k(x - c)^k\), where \(c\) is the center of the series.
Convergence is a major consideration when working with power series. We want to know where the series 'works,' that is, for which values of \(x\) it converges to a finite number. This is described by an interval or a radius of convergence. The key to finding this interval lies in applying certain mathematical tests.
In the example, each term of the series \(\frac{1}{k} x^{k}\) gets smaller as \(k\) increases, provided that \(x\) is within a certain range, which we are determining. If \(x\) is too large, the term won’t approach zero and the series may diverge.
Convergence is a major consideration when working with power series. We want to know where the series 'works,' that is, for which values of \(x\) it converges to a finite number. This is described by an interval or a radius of convergence. The key to finding this interval lies in applying certain mathematical tests.
In the example, each term of the series \(\frac{1}{k} x^{k}\) gets smaller as \(k\) increases, provided that \(x\) is within a certain range, which we are determining. If \(x\) is too large, the term won’t approach zero and the series may diverge.
Ratio Test
The Ratio Test is a handy tool for determining the convergence of a series, especially when you’re dealing with a power series. This test investigates the limit of the ratio of absolute terms in the sequence. Clearly, if terms are not getting smaller, the series may not converge.
For the Ratio Test, we express the test as \(L = \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\). If this limit \(L\) is less than 1, the series converges. If \(L\) is greater than 1, the series diverges. And if \(L\) equals 1, the test is inconclusive.
In the provided example, the Ratio Test is applied by taking the limit of \(\left|\frac{a_{k+1}}{a_k}\right|\) as \(k\) goes to infinity. Simplifying that ratio leaves us with the expression \(\left|\frac{k}{k+1}x\right|\), and as \(k\) becomes very large, the fraction approaches 1, leaving us with \(L = |x|\). By establishing that \(L < 1\), we have effectively used the Ratio Test to reveal information about the convergence of the power series.
For the Ratio Test, we express the test as \(L = \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|\). If this limit \(L\) is less than 1, the series converges. If \(L\) is greater than 1, the series diverges. And if \(L\) equals 1, the test is inconclusive.
In the provided example, the Ratio Test is applied by taking the limit of \(\left|\frac{a_{k+1}}{a_k}\right|\) as \(k\) goes to infinity. Simplifying that ratio leaves us with the expression \(\left|\frac{k}{k+1}x\right|\), and as \(k\) becomes very large, the fraction approaches 1, leaving us with \(L = |x|\). By establishing that \(L < 1\), we have effectively used the Ratio Test to reveal information about the convergence of the power series.
Limit of a Sequence
In mathematics, the concept of a limit is fundamental. It’s the value that a sequence or function 'approaches' as the index or input grows without bounds. For a sequence \(\{a_k\}\), the limit is written as \(\lim_{k \to \infty} a_k\) and can be understood as the value the sequence ‘settles into’ given enough terms.
Finding limits, especially for sequences that describe the terms of a power series, is a key step in determining a series’ behavior at the edges of its interval of convergence. As seen in our example, finding the limit as \(k\) approaches infinity for the ratio \(\frac{k}{k+1}\) is crucial. It tells us how the size of the terms in the series are changing—in this case, once simplified, it confirms that the terms do get smaller and hence support the idea of convergence as long as \(|x| < 1\).
Understanding the behavior of sequences and their limits not only aids in determining convergence but also provides insight into the nature of the functions that the power series represent.
Finding limits, especially for sequences that describe the terms of a power series, is a key step in determining a series’ behavior at the edges of its interval of convergence. As seen in our example, finding the limit as \(k\) approaches infinity for the ratio \(\frac{k}{k+1}\) is crucial. It tells us how the size of the terms in the series are changing—in this case, once simplified, it confirms that the terms do get smaller and hence support the idea of convergence as long as \(|x| < 1\).
Understanding the behavior of sequences and their limits not only aids in determining convergence but also provides insight into the nature of the functions that the power series represent.
Other exercises in this chapter
Problem 3
Find the interval of convergence. $$\sum k x^{k}$$
View solution Problem 4
Expand \(f(x)\) in powers of \(x,\) basing your calculations on the geometric series $$\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$ $$f(x)=\ln (1-x)$$
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Find the Taylor polynomial of the function \(f\) for the given values of \(a\) and \(n\) and give the Lagrange form of the remainder. $$f(x)=\ln x ; \quad a=1,
View solution Problem 4
Test these series for (a) absolute convergence, (b) conditional convergence. \(\frac{1}{2 \ln 2}-\frac{1}{3 \ln 3}+\frac{1}{4 \ln 4}-\frac{1}{5 \ln 5}+\cdots+(-
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