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Chapter 9: Problem 15

Find the radius of convergence. $$\sum_{n=0}^{\infty} \frac{(n+1) x^{n}}{2^{n}+n}$$

Short Answer

Expert verified
The radius of convergence is 2.

Step by step solution

01

Identify the general term

Consider the general term \( a_n = \frac{(n+1)x^n}{2^n+n} \) of the series.
02

Apply the Ratio Test

For the Ratio Test, calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). We find that \( a_{n+1} = \frac{(n+2)x^{n+1}}{2^{n+1} + (n+1)} \).
03

Simplify the Ratio Expression

Simplify \( \frac{a_{n+1}}{a_n} \) to get:\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| x \right| \cdot \frac{(n+2)}{(n+1)} \cdot \frac{2^n + n}{2^{n+1} + (n+1)} \]
04

Evaluate the Limit for the Ratio Test

We need to calculate:\[ \lim_{n \to \infty} \left| x \right| \cdot \frac{(n+2)}{(n+1)} \cdot \frac{2^n + n}{2^{n+1} + (n+1)} \]For large \(n\), the dominant terms are \(2^n\) in both the numerator and denominator:\[ \lim_{n \to \infty} \left| x \right| \cdot \frac{(n+2)}{(n+1)} \cdot \frac{2^n}{2^{n+1}} = \lim_{n \to \infty} \frac{1}{2} \left| x \right| \]Hence, the limit is \( \frac{1}{2} \left| x \right| \).
05

Determine the Radius of Convergence

The series converges when the limit \( \frac{1}{2} \left| x \right| < 1 \), which simplifies to \( \left| x \right| < 2 \). Therefore, the radius of convergence is \( R = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a crucial tool for determining the convergence of infinite series, especially power series. It's based on evaluating the limit of the ratio of successive terms in a series.
  • To use the Ratio Test, you find the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), where \( a_n \) is the general term of your series.
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or \( L = \infty \), the series diverges.
  • If \( L = 1 \), the test is inconclusive; other methods must be used.
In the context of power series, the Ratio Test helps to identify the radius of convergence, which is the range of values for which the series converges.The power of this test lies in its ability to simplify complex expressions and focus on the behavior of series as they approach infinity. This makes it much easier to determine where series converge.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where \( c_n \) is a sequence of coefficients and \( a \) is the center of the series.
  • Each term in a power series is a power of \( (x - a) \), adjusted by the coefficient \( c_n \).
  • The concept of power series is deeply linked with functions, as many functions can be expressed as power series within a certain interval.
  • The interval where a power series converges is crucial and is centered around the point \( a \), known as the center.
  • The distance from \( a \) to the endpoints of this interval is described by the radius of convergence \( R \).
Power series are powerful in calculus and analysis because they allow functions to be expressed in ways that are easier to differentiate and integrate than their original forms. Finding the series' radius of convergence is a critical part of understanding where this representation is valid.
Convergence
Convergence in the context of series refers to whether the sum of an infinite series approaches a finite value. If a series converges, you can sum its infinite terms to approximate a number.
  • Convergent series are those whose sequence of partial sums converge to a finite number.
  • Divergent series, on the other hand, do not sum to a finite value. They may increase indefinitely or oscillate.
  • Understanding convergence is critical as it tells us where a function, expressed as an infinite series, actually behaves predictably.
  • To find where a power series converges, the radius of convergence must be determined, which specifies the interval of \( x \)-values for convergence.
Several tests for convergence exist, including the Ratio Test, Root Test, and Integral Test, among others. Each is useful depending on the type of series you are dealing with. In our example, the Ratio Test provides an efficient means to measure convergence for the series at hand.
Limit Evaluation
Evaluating limits is a fundamental technique, especially in the context of the Ratio Test. It involves finding the value that a function or sequence "approaches" as the input approaches some point.
  • In series analysis, evaluating limits often involves simplifying complex expressions to see how they behave as \( n \to \infty \).
  • Proper limit evaluation helps in determining convergence, especially when applying tests like the Ratio Test.
  • In our example, simplifying \( \lim_{n \to \infty} \frac{1}{2} \left| x \right| \) helped identify the convergence behavior of the series.
  • Numerical and algebraic techniques are commonly used to perform these evaluations, focusing on the leading terms that dominate the behavior as \( n \) becomes very large.
Understanding how to evaluate limits in mathematical series can help break down complex expressions into more manageable parts, guiding the determination of convergence and other properties. It often involves focusing on dominant terms such as \( 2^n \) in our example, allowing simplification and limit calculation.

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Most popular questions from this chapter

Let \(\sum a_{n}\) be a series that converges to a number \(L\) and whose terms are all positive. We show that any rearrangement \(\sum c_{n}\) of this series also converges to \(L\) (a) Show that \(S_{n}K\) (c) Use parts (a) and (b) to show \(\sum c_{n}\) converges to \(L\)

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{n 2^{n}}{3^{n}}$$

Decide if the statements are true or false. Give an explanation for your answer. If all terms \(s_{n}\) of a sequence are less than a million, then the sequence is bounded.

Determine whether the series converges. $$\sum_{n=2}^{\infty} \frac{n+2}{n^{2}-1}$$

Decide if the statements are true or false. Give an explanation for your answer. The series \(\sum_{n=0}^{\infty}(-1)^{n} 2^{n}\) converges.
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