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Magazine Chapter 11: Problem 5
Find the radius of convergence and interval of convergence of the series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{2 n-1} $$
Short Answer
Expert verified
The radius of convergence is 1, and the interval of convergence is \((-1, 1]\).
Step by step solution
01
Identify the form of the series
The given series is \( \sum_{n=1}^{\infty} \frac{x^{n}}{2n-1} \). This series is in the power series form \( \sum_{n=1}^{\infty} a_n x^n \) where \(a_n = \frac{1}{2n-1}\).
02
Apply the ratio test
We use the ratio test to find the radius of convergence. The ratio test involves finding \(|a_{n+1} x^{n+1} / a_n x^n|\). Here, \(a_{n+1} = \frac{1}{2(n+1)-1}\) and \(a_n = \frac{1}{2n-1}\).
03
Simplify the ratio for the limit
Compute the ratio: \[ \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = \left| \frac{x \cdot (2n-1)}{2n+1} \right|. \] We replace \( a_{n+1}, a_n \) into their expressions and simplify.
04
Calculate the limit as \(n\) approaches infinity
To find the radius of convergence \(R\), compute the limit \(L = \lim_{n \to \infty} |x| \cdot \frac{2n-1}{2n+1}\), which simplifies to \( |x| \). Thus, for convergence, we need \( |x| < R = 1 \). Hence, \( R = 1 \).
05
Determine the interval of convergence
With \( R = 1 \), initially, the interval is \(-1 < x < 1\). We need to check endpoints \(x = -1\) and \(x = 1\) separately to see if the series converges at these points.
06
Evaluate convergence at \(x = 1\)
If \(x = 1\), the series becomes \(\sum_{n=1}^{\infty} \frac{1}{2n-1}\), which is the harmonic series of odd terms. It diverges.
07
Evaluate convergence at \(x = -1\)
If \(x = -1\), the series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\), which is an alternating series. This series converges by the alternating series test.
08
Conclude the interval of convergence
Since the series converges only at \(x = -1\) and not at \(x = 1\), the interval of convergence is \((-1, 1]\).
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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 a series of the form \( \sum_{n=0}^{\infty} a_n x^n \), where \(a_n\) is the coefficient of the series, \(x\) is a variable, and \(n\) is a non-negative integer.
Power series are essential in calculus and mathematical analysis because they can represent functions as infinite polynomials within a range of convergence.
Power series are essential in calculus and mathematical analysis because they can represent functions as infinite polynomials within a range of convergence.
- Power series often depend on a variable \(x\) that can be any real or complex number.
- Each power series has a center, often at \(x = 0\) (known as a Maclaurin series) or at some other point \(x = c\) (known as a Taylor series).
- The behavior of a power series largely revolves around the concepts of convergence and divergence.
Interval of Convergence
The interval of convergence of a power series is the set of all values of \(x\) for which the series converges to a finite value.
It includes all the open values within the bounds and may include some or none of the endpoints.To determine the interval of convergence, we need:
However, checking the endpoints one by one reveals that:
It includes all the open values within the bounds and may include some or none of the endpoints.To determine the interval of convergence, we need:
- A radius of convergence, denoted as \(R\).
- Verification of convergence at the endpoints of the interval.
However, checking the endpoints one by one reveals that:
- At \(x = 1\), it diverges.
- At \(x = -1\), the series converges by the alternating series test.
Ratio Test
The ratio test is a method used to determine the absolute convergence of an infinite series.
It states that for a series \( \sum a_n \), the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) determines convergence.
Compute \[L = \lim_{n \to \infty} \left| \frac{x \cdot (2n-1)}{2n+1} \right|\]This simplifies to \(|x|\), giving the radius of convergence as \( R = 1 \).
The test helps establish the primary interval \(-1 < x < 1\) before checking endpoints.
It states that for a series \( \sum a_n \), the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) determines convergence.
- If \(L < 1\), the series converges absolutely.
- If \(L > 1\) or if \(L\) is infinite, the series diverges.
- If \(L = 1\), the test is inconclusive; other methods must be used.
Compute \[L = \lim_{n \to \infty} \left| \frac{x \cdot (2n-1)}{2n+1} \right|\]This simplifies to \(|x|\), giving the radius of convergence as \( R = 1 \).
The test helps establish the primary interval \(-1 < x < 1\) before checking endpoints.
Alternating Series Test
The alternating series test is a handy method used to verify the convergence of series where successive terms alternate in sign.
This test is helpful for series of the form \( \sum (-1)^n a_n \), where the sequence \( a_n \) is positive, decreasing, and approaches zero.
This test is helpful for series of the form \( \sum (-1)^n a_n \), where the sequence \( a_n \) is positive, decreasing, and approaches zero.
- The test states: if the terms \( a_n \) decrease steadily to zero, then the series converges.
- The terms \( \frac{1}{2n-1} \) are positive and decrease to zero.
- By the alternating series test, this series converges at that endpoint.
