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Use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty}(-1)^{n} n^{2} e^{-n}$$

Short Answer

Expert verified
The series converges by the Alternating Series Test.

Step by step solution

01

Identify the Series

The given series is \( \sum_{n=1}^{\infty} (-1)^{n} n^{2} e^{-n} \). This is an alternating series because the term \( (-1)^{n} \) alternates the sign of each subsequent term.
02

Check for Alternating Series Test

For an alternating series to converge, two conditions must be satisfied: (1) the absolute value of the terms \( a_n = n^2 e^{-n} \) must decrease as \( n \) increases, and (2) \( \lim_{{n \to \infty}} a_n = 0 \).
03

Verify Decreasing Condition

To confirm that \( a_n = n^2 e^{-n} \) is decreasing, observe the derivative. This involves finding the derivative of \( f(n) = n^2 e^{-n} \) and showing it is negative for all \( n \geq 1 \). This can be calculated, or we can argue intuitively that since \( e^{-n} \) decreases exponentially faster than \( n^2 \) increases, the product \( n^2 e^{-n} \) decreases.
04

Evaluate Limit as n Approaches Infinity

Calculate \( \lim_{{n \to \infty}} n^2 e^{-n} \). Notice that as \( n \) approaches infinity, \( e^{-n} \) approaches zero much more rapidly than \( n^2 \) increases. Thus, \( \lim_{{n \to \infty}} n^2 e^{-n} = 0 \).
05

Conclusion

Since \( a_n = n^2 e^{-n} \) is decreasing and approaches zero as \( n \) approaches infinity, the series \( \sum_{n=1}^{\infty} (-1)^{n} n^2 e^{-n} \) converges by the Alternating Series Test.

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Key Concepts

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

Convergence Tests
Convergence tests are essential tools in determining whether an infinite series converges or diverges. Think of them as different methods or lenses to examine a series. The alternating series test is just one type, but others include the ratio test, root test, and integral test. Each test provides a mechanism to assess the behavior of series, particularly as the number of terms increases towards infinity. Convergence is crucial because it tells us if the sum of an infinite series settles on a particular number rather than growing indefinitely.

If a series converges, the sum of its infinite terms results in a finite number. The process often involves testing conditions related to the terms of the series, such as their order, sign, and magnitude. It's important for students to become familiar with different tests and understand which situations call for which test.
Absolute Terms Decreasing
For the alternating series test to apply, a key condition is that the absolute terms of the series must be decreasing. This means that each term is smaller in magnitude than the previous term. In mathematical terms, if a sequence of terms is given by \(a_n\), then for every \(n\), it must hold that \(a_{n+1} \leq a_n\).

This requirement ensures that the impact of each new term is smaller than the one before it, preventing the series from oscillating too wildly and potentially diverging. Decreasing absolute terms help stabilize the series towards convergence. In our exercise, even though the term \(n^2\) grows, \(e^{-n}\) diminishes rapidly enough to outweigh this, thus the terms \(a_n = n^2 e^{-n}\) decrease for growing \(n\).

Checking whether terms are decreasing can involve taking derivatives or comparing successive terms.
Calculus Series
In calculus, series represent the summing of sequences—a fundamental concept with vast applications. Specifically, a series can be thought of as adding terms from a sequence infinitely. The calculation of sums becomes complex when dealing with an infinite number of terms.

A keen grasp of series helps in understanding functions, especially in approximations and solutions to differential equations. Series in calculus, such as power series, Fourier series, and Taylor series, allow mathematicians and scientists to express functions in simplified or approximate forms. Understanding convergence is key because it determines if the infinite sum accurately represents something meaningful or understandable.

In our specific series \(\sum_{n=1}^{\infty} (-1)^{n} n^2 e^{-n}\), the alternating sign introduces a unique consideration as opposed to series with positive terms only.
Alternating Series Test
The alternating series test is a specific convergence test used for series whose terms alternate in sign. It provides a straightforward criterion for convergence:
  • The absolute values of the terms must be decreasing.
  • The limit of the absolute term as \(n\) approaches infinity must be zero, \(\lim_{{n \to \infty}} a_n = 0\).
If both conditions are met, the series converges.

In our exercise, the series \(\sum_{n=1}^{\infty} (-1)^{n} n^2 e^{-n}\) fulfills these conditions. The sequence \(n^2 e^{-n}\) meets the first condition by decreasing, and for the second condition, its limit equals zero. Thus, the series converges by the alternating series test. This test is incredibly useful for such types of series as it shields the complexity sometimes involved in other convergence tests while still offering a reliable conclusion.

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Most popular questions from this chapter

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

The English mathematician Wallis discovered the formula $$\frac{\pi}{4}=\frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot \cdots}$$ Find \(\pi\) to two decimal places with this formula.

Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=\frac{4^{n+1}+3^{n}}{4^{n}}\)

a. Suppose that \(f(x)\) is differentiable for all \(x\) in [0,1] and that \(f(0)=0 .\) Define sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=n f(1 / n)\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0) .\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\) \(\mathbf{b} \cdot a_{n}=n \tan ^{-1} \frac{1}{n} \quad\) c. \(a_{n}=n\left(e^{1 / n}-1\right)\) d. \(a_{n}=n \ln \left(1+\frac{2}{n}\right)\)

A sequence of rational numbers is described as follows: \(\frac{1}{1}, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \ldots, \frac{a}{b}, \frac{a+2 b}{a+b}, \ldots\) Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let \(x_{n}\) and \(y_{n}\) be, respectively, the numerator and the denominator of the \(n\) th fraction \(r_{n}=x_{n} / y_{n}\) a. Verify that \(x_{1}^{2}-2 y_{1}^{2}=-1, x_{2}^{2}-2 y_{2}^{2}=+1\) and, more generally, that if \(a^{2}-2 b^{2}=-1\) or \(+1,\) then \((a+2 b)^{2}-2(a+b)^{2}=+1 \quad\) or \(\quad-1\) respectively. b. The fractions \(r_{n}=x_{n} / y_{n}\) approach a limit as \(n\) increases. What is that limit? (Hint: Use part (a) to show that \(\left.r_{n}^{2}-2=\pm\left(1 / y_{n}\right)^{2} \text { and that } y_{n} \text { is not less than } n .\right)\)

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