Chapter 9: Problem 26
Show that the series \(\sum_{n=1}^{\infty} \frac{(2 x)^{n}}{n}\) converges for \(|x|<1 / 2\) Investigate whether the series converges for \(x=1 / 2\) and \(x=-1 / 2.\)
Short Answer
Expert verified
The series converges for \(|x|<1/2\). It diverges at \(x=1/2\) and converges at \(x=-1/2\).
Step by step solution
01
Analyze the series
The given series is \(\sum_{n=1}^{\infty} \frac{(2x)^n}{n}\). This is a power series in the form \(\sum_{n=1}^{\infty} c_n x^n\) with coefficients \(c_n = \frac{2^n}{n}\). We are tasked to find its interval of convergence.
02
Use the Ratio Test for Convergence
To determine the interval of convergence, apply the Ratio Test: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(2x)^{n+1}}{n+1} \cdot \frac{n}{(2x)^n} \right| = \lim_{n \to \infty} \left| \frac{2nx}{n+1} \right| = |2x| \] For convergence, we need \(|2x| < 1\), which simplifies to \(|x| < 1/2\).
03
Investigate Edge Cases \(x = 1/2\) and \(x = -1/2\)
Substitute \(x = 1/2\) into the series: \[ \sum_{n=1}^{\infty} \frac{(2 \cdot 1/2)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n} \] This is the harmonic series, which diverges. For \(x = -1/2\): \[ \sum_{n=1}^{\infty} \frac{(2 \cdot -1/2)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \] This is the alternating harmonic series, which converges.
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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 \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where * \( c_n \) are the coefficients of the series, * \( x \) is the variable, and * \( a \) is the center around which the series is expanded.This series functions like a polynomial, but with infinitely many terms.Power series are significant in many fields like physics and engineering because they allow complex functions to be expressed in simpler polynomial-like terms. Each value of \( x \) within a certain range, called the interval of convergence, makes the series sum up to a finite value. Understanding the interval of convergence is crucial because it tells us for which values of \( x \) the series will work without diverging towards infinity.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of a series. For a given series \( \sum a_n \), the Ratio Test involves calculating the limit \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]This limit helps us decide the nature of the series:- If \( L < 1 \), the series converges absolutely.- If \( L > 1 \) or is infinite, the series diverges.- If \( L = 1 \), the test is inconclusive, and other methods need to be used.For example, when applying the Ratio Test to the power series \( \sum_{n=1}^{\infty} \frac{(2x)^n}{n} \), we found \[ L = |2x|\]For the series to converge, \(|2x| < 1\), which simplifies to \(|x| < 1/2\). This determines the basic interval initially, but further checks for end points are necessary.
Interval of Convergence
The interval of convergence of a power series is the range of \( x \) values for which the series converges. It is crucial for understanding where a series behaves nicely and gives finite results.To find the interval, we often use tests like the Ratio Test or the Root Test. The endpoints require separate checks because sometimes a series can converge at one endpoint and not the other.In the given series \( \sum_{n=1}^{\infty} \frac{(2x)^n}{n} \), the Ratio Test provides that the interval is \(|x| < 1/2\).We further explored this:* At \( x = 1/2 \), the series resembles a harmonic series, known to diverge.* At \( x = -1/2 \), it becomes an alternating series that converges.Thus the true interval is \((-1/2, 1/2]\). Checking these boundaries ensures that we understand the series thoroughly.
Alternating Series
An alternating series is a series whose terms alternate in sign, such as \( \sum (-1)^n a_n \). These series have their own convergence criteria, primarily governed by the Alternating Series Test, which requires:- The absolute value of the terms \( a_n \) decreases continuously.- The limit of \( a_n \) as \( n \) approaches infinity is 0.The famous alternating series is the alternating harmonic series, \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), which converges despite its non-alternating counterpart diverging.In our example, at \( x = -1/2 \), the series becomes an alternating harmonic series. Each term's sign alternates due to \( (-1)^n \), allowing the series to meet all the convergence conditions.
Harmonic Series
The harmonic series is one of the most famous examples of a divergent series. It is written as \( \sum_{n=1}^{\infty} \frac{1}{n} \) and despite its seemingly decreasing terms, it diverges to infinity.This series is interesting because its divergence shows that even a series with diminishing terms can add up to infinity.In our exercise, when \( x = 1/2 \), the power series transmuted into a harmonic series, highlighted as:\[ \sum_{n=1}^{\infty} \frac{1}{n} \]This transformation demonstrates that at specific values, a benign-looking power series can yield a divergent harmonic series, reminding us of the intricacies involved in convergence analysis.