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Chapter 11: Problem 8

Test the series for convergence or divergence. $$\sum_{k=1}^{\infty} \frac{2^{k} k !}{(k+2) !}$$

Short Answer

Expert verified
The series diverges by the Ratio Test.

Step by step solution

01

Identify Series Type

We are given the series \( \sum_{k=1}^{\infty} \frac{2^{k} k!}{(k+2)!} \). To determine convergence, we need to find an appropriate test to apply. The series resembles a ratio or factorial series, so the Ratio Test is a good option to try.
02

Apply Ratio Test

To use the Ratio Test, consider the general term \( a_k = \frac{2^{k} k!}{(k+2)!} \). We find the limit of \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Compute \( a_{k+1} = \frac{2^{k+1} (k+1)!}{(k+3)!} \).
03

Simplify Terms in Ratio

Now compute the Ratio: \( \frac{a_{k+1}}{a_k} = \frac{2^{k+1} (k+1)! / (k+3)!}{2^k k! / (k+2)!} = \frac{2 \cdot (k+1)! \cdot (k+2)!}{k! \cdot (k+3)!} \). Simplifying gives: \( \frac{2 \cdot (k+1) \cdot (k+2)}{(k+3)} \).
04

Evaluate the Limit

Now find the limit: \( \lim_{k \to \infty} \frac{2(k+1)(k+2)}{(k+3)} \). This simplifies to \( \lim_{k \to \infty} \frac{2(k^2 + 3k + 2)}{k+3} \). As \( k \to \infty \), the dominant term in the numerator and denominator is \( k^2 \), so the expression simplifies to \( 2k \), which leads to \( \lim_{k \to \infty} 2 \).
05

Conclude with Ratio Test Result

The limit \( 2 \) is greater than \( 1 \), according to the Ratio Test. If the limit \( L > 1 \), the series diverges. Hence, the given series diverges.

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Key Concepts

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

Convergence
Convergence in the context of infinite series refers to a series whose terms approach a specific value as more terms are added. When an infinite series converges, the sum of its terms becomes closer and closer to a definite number. This implies that, as we keep summing the terms, the series gets nearer to a particular limit, which we call the series limit.
  • A convergent series implies stability in growth, meaning it doesn't grow infinitely.
  • The series will typically consist of terms that get smaller and eventually tend to zero.
  • In mathematical analysis, convergence is a core concept for determining the behavior of series.
To establish convergence for any series, tests like the n-th term test, Ratio Test, or Integral Test are commonly used. Only when a series passes these tests can we confirm its convergence.
Divergence
Divergence is essentially the opposite of convergence. In a divergent series, the sum of the series either heads towards infinity or fluctuates without settling at a finite value. In simpler terms, if you keep adding the numbers in a divergent series, you'll never reach a stable sum.
  • A divergent series is commonly associated with terms that do not diminish to zero.
  • The sum of such series increases or fails to stabilize as terms are added.
  • Tests like the Ratio Test can highlight when a series diverges.
In the given exercise, the series diverges because, when using the Ratio Test, the limit resulted in a number greater than one, indicating endless, unbounded growth rather than settling at a finite number.
Factorial Series
A factorial series involves terms that contain factorials, like \( k! \) or \( (k+n)! \). These grow very fast because a factorial means multiplying a sequence of descending positive integers. For example, \( 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \). The function grows rapidly as \( k\) increases.
  • Factorials can create series that are initially small in magnitude but grow quite large.
  • This quality means factorial series require specific tests for convergence or divergence, with the Ratio Test being a popular choice.
  • Factorials in the denominator (such as in the given series) tend to hamper growth, offering more potential for convergence.
Understanding factorials is critical when working with series like the one in the exercise. Despite having factorials in both numerator and denominator, the provided series still diverges when applying the Ratio Test.
Infinite Series
An infinite series is a sum of terms that continues without end. In mathematics, infinite series can either converge or diverge, and each has profound implications in calculus and numerical analysis.
  • They can be expressed in general terms like \( \sum_{k=1}^{\infty} a_k \), indicating a series that starts at a certain point and extends indefinitely.
  • Infinite series are central to calculus concepts, from basic functions to complex integrals.
  • Analysis of such series often involves determining convergence or divergence, using a toolset that includes various tests and criteria.
The series in the original problem is an example of an infinite series. Despite its construction using factorials and exponential terms, the goal is always to assess its behavior at infinity to determine its convergence or divergence, using techniques like the Ratio Test.

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Most popular questions from this chapter

Show that the sequence defined by $$ a_{1}=2 \quad a_{n+1}=\frac{1}{3-a_{n}} $$ satisfies \(0< a_{n} \leqslant 2\) and is decreasing. Deduce that the sequence is convergent and find its limit.

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \frac{\sqrt{n^{2}-1}}{n^{3}+2 n^{2}+5}$$

A function \(f\) is defined by $$f(x)=1+2 x+x^{2}+2 x^{3}+x^{4}+\cdots$$ that is, its coefficients are \(c_{2 n}=1\) and \(c_{2 n+1}=2\) for all \(n \geqslant 0\) . Find the interval of convergence of the series and find an explicit formula for \(f(x)\) that is, its coefficients are \(c_{2 n}=1\) and \(c_{2 n+1}=2\) for all \(n \geqslant 0\) . Find the interval of convergence of the series and find an explicit formula for \(f(x) .\)

Find the Maclaurin series of \(f\) (by any method) and its radius of convergence. Graph \(f\) and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and \(f\) ? \(f(x)=\ln \left(1+x^{2}\right)\)

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