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Chapter 8: Problem 8

Find the radius of convergence and interval of conver- gence of the series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n 3^{n}} $$

Short Answer

Expert verified
The radius of convergence is 3, and the interval of convergence is [-3, 3).

Step by step solution

01

Identify the Series Form

The given series is \( \sum_{n=1}^{\infty} \frac{x^n}{n \, 3^n} \). This is a power series of the form \( \sum_{n=1}^{\infty} a_n x^n \) where \( a_n = \frac{1}{n \, 3^n} \).
02

Apply the Ratio Test for Convergence

The Ratio Test is used to determine the convergence of a power series by examining the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). For our series, this is \( L = \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1) \, 3^{n+1}} \times \frac{n \, 3^n}{x^n} \right| \). Simplifying gives \( \left| \frac{x}{n+1} \times \frac{n}{3} \right| = \left| \frac{x}{3} \right| \times \frac{n}{n+1} \).
03

Simplify the Ratio and Calculate the Limit

The limit simplifies as \( \lim_{n \to \infty} \left| \frac{x}{3} \right| \times \frac{n}{n+1} = \left| \frac{x}{3} \right| \) because \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \).
04

Determine the Radius of Convergence

The series converges if \( L < 1 \). Thus, we have \( \left| \frac{x}{3} \right| < 1 \), leading to \( \left| x \right| < 3 \). Therefore, the radius of convergence \( R \) is 3.
05

Identify the Interval of Convergence

The interval of convergence before checking endpoints is \( -3 < x < 3 \). We must check the endpoints \( x = -3 \) and \( x = 3 \) separately.
06

Check Convergence at Endpoints

At \( x = -3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-3)^n}{n \, 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), which is the alternating harmonic series and converges.At \( x = 3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{3^n}{n \, 3^n} = \sum_{n=1}^{\infty} \frac{1}{n} \), which diverges.
07

Write the Final Interval of Convergence

Since the series converges at \( x = -3 \) and diverges at \( x = 3 \), the interval of convergence is \( [-3, 3) \).

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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 type of infinite series that takes the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \). It鈥檚 a method of representing functions in terms of variable powers. In this formula, \( a_n \) are the coefficients, \( c \) is the center of the series, and \( x \) is the variable.
  • Power series can represent functions over a certain range of \( x \) values.
  • The series converges depending on the values of \( x \) relative to the center \( c \).
  • Determining the radius of convergence helps identify the range of \( x \) where the series is valid.
This form offers a way to approach complex functions by expressing them in terms of simpler polynomial approximations.Breaking down a complex function into a sum of its powers simplifies analysis and calculations.
Ratio Test
The Ratio Test is a method used to determine the convergence of a series, particularly useful for power series. It involves evaluating the limit:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or is infinite, the series diverges.
  • If \( L = 1 \), the test is inconclusive.
This process is useful because it tells us how the series behaves as \( n \) goes to infinity. When applied to our exercise, this test helps to find the radius of convergence by simplifying the limit expression and identifying the conditions under which the series converges. It helps ease the complexity by systematically reducing the series to a simple term form to test convergence.
Interval of Convergence
The interval of convergence is the set of \( x \) values for which a power series converges. Once the radius of convergence is determined, we can define the interval:
  • For a series centered at \( c \) with radius \( R \), the basic interval is \((c-R, c+R)\).
  • Endpoints \( c-R \) and \( c+R \) need to be checked separately to see if the series converges at these points.
In our example, with a radius of convergence of 3, the initial interval is \(-3 < x < 3\). Checking endpoints is crucial because convergence can vary at these critical points, impacting the final interval. Analyzing this ensures completeness in the solution by including or excluding endpoints based on their convergence behavior. This thorough analysis confirms the exact domain where the series holds true.
Convergence Tests
Convergence tests are essential tools that help determine whether a series converges or diverges. Various tests exist, each applicable under different circumstances:
  • Ratio Test: As mentioned, effective for power series.
  • Root Test: Considers the nth root of terms, useful for exponential terms.
  • Alternating Series Test: Applied to series whose terms alternate in sign, checking for conditional convergence.
By selecting the appropriate test, you can efficiently decide the convergence behavior of the series. For our specific series, the Ratio Test gave a clear understanding of the radius and check of convergence, leading to insightful conclusions about the interval of convergence. Using convergence tests ensures rigorous analysis, confirming the series' behavior over its domain. They simplify the decision-making process when determining convergence properties of different series.

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Most popular questions from this chapter

(a) Show that the function $$ f(x)=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$ is a solution of the differential equation $$ f^{\prime}(x)=f(x) $$ (b) Show that \(f(x)=e^{x}\)

(a) A sequence \(\left\\{a_{n}\right\\}\) is defined recursively by the equation \(a_{n}=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)\) for \(n \geqslant 3,\) where \(a_{1}\) and \(a_{2}\) can be any real numbers. Experiment with various values of \(a_{1}\) and \(a_{2}\) and use your calculator to guess the limit of the sequence. (b) Find \(\lim _{n \rightarrow \infty} a_{n}\) in terms of \(a_{1}\) and \(a_{2}\) by expressing \(a_{n+1}-a_{n}\) in terms of \(a_{2}-a_{1}\) and summing a series.

If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth. (a) If \(R\) is the radius of the earth and \(L\) is the length of the highway, show that the correction is $$C=R \sec (L / R)-R$$ (b) Use a Taylor polynomial to show that $$C \approx \frac{L^{2}}{2 R}+\frac{5 L^{4}}{24 R^{3}}$$ Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 \(\mathrm{km}\) long. (Take the radius of the earth to be 6370 \(\mathrm{km.}\) )

If the \(n\) th partial sum of a series \(\sum_{n=1}^{\infty} a_{n}\) is $$s_{n}=\frac{n-1}{n+1}$$ find \(a_{n}\) and \(\sum_{n=1}^{\infty} a_{n}\)

We have seen that the harmonic series is a divergent series whose terms approach 0. Show that $$\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right)$$ is another series with this property.
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