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Chapter 10: Problem 40

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \operatorname{sech}^{2} n $$

Short Answer

Expert verified
The series converges by the limit comparison test with a geometric series.

Step by step solution

01

Understanding the Series

The given series is \( \sum_{n=1}^{\infty} \operatorname{sech}^{2} n \), where \( \operatorname{sech}(n) \) is the hyperbolic secant of \( n \), equivalent to \( \frac{1}{\cosh(n)} \). The term \( \operatorname{sech}^{2}(n) \) simplifies to \( \frac{1}{\cosh^2(n)} \) for each integer \( n \geq 1 \).
02

Check for Convergence Using the Limit Comparison Test

Consider comparing it to a known convergent series. Since \( \cosh(n) = \frac{e^n + e^{-n}}{2} \), as \( n \to \infty \), \( \cosh(n) \approx \frac{e^n}{2} \), so \( \operatorname{sech}(n) \approx 2 e^{-n} \) and \( \operatorname{sech}^2(n) \approx 4 e^{-2n} \). Compare it to the geometric series \( \sum 4 e^{-2n} \).
03

Analyze the Geometric Series

The geometric series \( \sum_{n=1}^{\infty} 4 e^{-2n} \) can be rewritten as \( 4 \sum_{n=1}^{\infty} (e^{-2})^n \). It converges because it is a geometric series with a common ratio \( r = e^{-2} \), where \( 0 < r < 1 \).
04

Apply the Comparison Test

Since \( \operatorname{sech}^2(n) \approx 4 e^{-2n} \) and the series \( \sum 4 e^{-2n} \) converges, the given series \( \sum \operatorname{sech}^2(n) \) should also converge by the limit comparison test. We find that \( \lim_{n \to \infty} \frac{\operatorname{sech}^2(n)}{4 e^{-2n}} = 1 \). Since the limit is finite and non-zero, the comparison test confirms convergence.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a tool used to determine whether a given series converges or diverges by comparing it with a second series. To use this test, you need another series whose convergence properties are known.

Here's how it works:
  • Identify a series you want to compare, say, \( b_n \), which has known convergence or divergence.
  • Calculate the limit \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \), where \( a_n \) are the terms of the series you're analyzing.
  • If the limit \( L \) is a positive, finite number, then both \( \sum a_n \) and \( \sum b_n \) either both converge or both diverge.
In our exercise, the series \( \sum \operatorname{sech}^2(n) \) is compared with the known geometric series \( \sum 4e^{-2n} \). By showing that \( \lim_{{n \to \infty}} \frac{\operatorname{sech}^2(n)}{4e^{-2n}} = 1 \), we confirm the series converges since \( \sum 4e^{-2n} \) is already known to converge.
Geometric Series
A geometric series is one of the most straightforward types of series to analyze in terms of convergence and divergence. It is a series of the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

Key points about geometric series:
  • If \( |r| < 1 \), the series converges. The sum is given by \( \frac{a}{1-r} \).
  • If \( |r| \geq 1 \), the series diverges.
In the original exercise, we look at the series \( \sum 4 e^{-2n} \), which can be rewritten as the geometric series \( 4 \sum (e^{-2})^n \). Here, the common ratio is \( r = e^{-2} \), which satisfies \( 0 < r < 1 \), thus confirming the series converges.
Hyperbolic Functions
Hyperbolic functions are analogues of the usual trigonometric functions but are based on hyperbolas rather than circles. They include functions like the hyperbolic sine \( \sinh(x) \), hyperbolic cosine \( \cosh(x) \), and hyperbolic secant \( \operatorname{sech}(x) \).

Understanding hyperbolic functions:
  • The hyperbolic cosine is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
  • The hyperbolic secant is the reciprocal of hyperbolic cosine, \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \).
  • For large values of \( n \), \( \cosh(n) \) approximates \( \frac{e^n}{2} \), making \( \operatorname{sech}(n) \approx 2e^{-n} \).
In our solution, recognizing \( \operatorname{sech}^2(n) \approx 4e^{-2n} \) allows us to apply convergence tests like the Limit Comparison Test effectively.

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Most popular questions from this chapter

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The Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is \(\sum_{n=0}^{\infty} a_{n} x^{n}\) A function defined by a power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) with a radius of convergence \(R>0\) has a Taylor series that converges to the function at every point of \((-R, R) .\) Show this by showing that the Taylor series generated by \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) is the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) itself. \begin{equation} \begin{array}{c}{\text { An immediate consequence of this is that series like }} \\ {x \sin x=x^{2}-\frac{x^{4}}{3 !}+\frac{x^{6}}{5 !}-\frac{x^{8}}{7 !}+\cdots} \\ {\text {and}}\end{array} \end{equation} \begin{equation} x^{2} e^{x}=x^{2}+x^{3}+\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\cdots \end{equation} obtained by multiplying Taylor series by powers of \(x,\) as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.

Uniqueness of convergent power series a. Show that if two power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(\sum_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) . (Hint: Let \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !) . )\) b. Show that if \(\sum_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n .\)
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