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Chapter 11: Problem 2

\(2-28\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}}$$

Short Answer

Expert verified
The series is absolutely convergent.

Step by step solution

01

Define Absolute Convergence

A series \( \sum a_n \) is absolutely convergent if the series \( \sum |a_n| \) is convergent. For our series, \( a_n = \frac{n^2}{2^n} \). Since \( a_n \) is already positive, \( |a_n| = a_n \). So, we need to determine if \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \) is convergent.
02

Apply the Ratio Test

The Ratio Test states that a series \( \sum a_n \) converges absolutely if \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). For \( a_n = \frac{n^2}{2^n} \), we have:\[\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 / 2^{n+1}}{n^2 / 2^n} = \frac{(n+1)^2}{2n^2}\]This simplifies to \( \frac{(n+1)^2}{2n^2} = \frac{n^2 + 2n + 1}{2n^2} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2} \).
03

Evaluate the Limit

Evaluate \( \lim_{n\to\infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2} \):\[\lim_{n\to\infty} \left( \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2} \right) = \frac{1 + 0 + 0}{2} = \frac{1}{2}\]Since \( \frac{1}{2} < 1 \), the series \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \) is absolutely convergent by the Ratio Test.

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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 powerful tool used to determine the convergence of a series. It specifically helps in assessing whether a series converges absolutely. This test involves taking the limit of the absolute value of the ratio of consecutive terms in the series.Here's how it works:
  • For the series \( \sum a_n \), calculate \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If the result of this limit is less than 1, the series is absolutely convergent.
  • If the result equals 1, the Ratio Test is inconclusive.
  • If the result is greater than 1, or if the limit does not exist, the series is divergent.

In the given problem, for \( a_n = \frac{n^2}{2^n} \), we calculated:\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^2/2^{n+1}}{n^2/2^n} = \frac{(n+1)^2}{2n^2} \]This simplifies to:\[ \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{2} \]As \( n \) approaches infinity, the limit is \( \frac{1}{2} \), confirming absolute convergence as \( \frac{1}{2} < 1 \).
Absolute Convergence
Absolute convergence of a series means that the series of absolute values converges, regardless of whether the original series does. If a series \( \sum a_n \) converges absolutely, then the series \( \sum |a_n| \) must converge as well.This is an important distinction because it ensures stronger criteria for convergence:
  • Absolute convergence implies both absolute and conditional convergence.
  • If \( \sum |a_n| \) converges, then \( \sum a_n \) also converges, but the reverse is not always true.

In the exercise, since \( a_n = \frac{n^2}{2^n} \) is positive for all terms, \( |a_n| = a_n \). We checked whether the series \( \sum \frac{n^2}{2^n} \) is convergent using the Ratio Test. With a result showing \( \frac{1}{2} < 1 \), it is confirmed to be absolutely convergent. Hence, both the original and absolute series converge.
Series Convergence
Series convergence refers to a series \( \sum a_n \) where the sum of its terms approaches a finite limit as the number of terms grows infinitely.There are different types of convergence:
  • Absolute Convergence: The series \( \sum |a_n| \) converges, which also implies the original series converges.
  • Conditional Convergence: The series \( \sum a_n \) converges but its absolute series \( \sum |a_n| \) does not.
  • Divergence: The series does not converge to a finite limit.

In our specific series, we determined it to be absolutely convergent through the Ratio Test. This means not only that \( \sum \frac{n^2}{2^n} \) converges, but also \( \sum \left| \frac{n^2}{2^n} \right| \), indicating robust convergence. Therefore, this series cannot be simply conditionally convergent.

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Most popular questions from this chapter

Suppose that the power series \(\Sigma c_{a}(x-a)^{n}\) satisfies \(c_{n} \neq 0\) for all \(n\) . Show that if \(\lim _{n \rightarrow \infty}\left|c_{n} / c_{n+1}\right|\) exists, then it is equal to the radius of convergence of the power series.

(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 \(f(x)=e^{x}\)

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function. \(y=e^{x} \ln (1-x)\)

\(27-30\) Use a power series to approximate the definite integral to six decimal places.\ $$ \int_{0}^{0.3} \frac{x^{2}}{1+x^{4}} d x $$

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function. \(y=\frac{x}{\sin x}\)
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