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Magazine Chapter 8: Problem 21
Use the Integral Test to determine the convergence or divergence of the following series, or state that the test does not apply. $$\sum_{k=1}^{\infty} k e^{-2 k^{2}}$$
Short Answer
Expert verified
Short Answer: The series \(\sum_{k=1}^{\infty} k e^{-2 k^{2}}\) converges by the Integral Test.
Step by step solution
01
Define the function and check conditions for the Integral Test
First, define a function \(f(k) = k e^{-2 k^{2}}\) which corresponds to the terms in the given series. Before applying the Integral Test, we need to check that \(f(k)\) is positive, continuous, and decreasing for \(k \ge 1\).
1. The function is positive for all \(k \ge 1\) since each term \(k\) and \(e^{-2k^2}\) is positive.
2. The function is continuous as the product of two continuous functions (\(k\) and \(e^{-2 k^{2}}\)).
3. To prove that the function is decreasing, let's find its derivative: \(f'(k) = (1 - 4k^2)e^{-2k^2}\). Since \(e^{-2k^2}\) is always positive for all \(k\), the sign of \(f'(k)\) is determined by the factor \((1 - 4k^2)\). For all \(k \ge 1\), \(-4k^2 \le -4 < 0\), so \(f'(k) \le 0\), which means the function is decreasing.
Since all conditions are satisfied, we can apply the Integral Test.
02
Apply the Integral Test
Now, we need to compute the improper integral of \(f(k)\) from 1 to infinity. By applying integration by parts:
Let \(u = k\) and \(dv = e^{-2k^2}dk\). Then, \(du = dk\) and \(v = -\frac{1}{2}e^{-2k^2}\). Apply integration by parts:
$$\int k e^{-2k^2}dk = -\frac{1}{2}ke^{-2k^2} - (-\frac{1}{2})\int e^{-2k^2}dk.$$
Now we substitute back the bounds of integration:
$$\int_1^{\infty} -\frac{1}{2}ke^{-2k^2}dk = \lim_{a\to\infty} (-\frac{1}{2}ae^{-2a^2} + \frac{1}{2}e^{-2} + \frac{1}{2}\int_1^a e^{-2k^2}dk).$$
As \(a\to\infty\), \(ae^{-2a^2}\) approaches 0 because the exponential term dominates, making the product approach 0. So we are left with the improper integral:
$$\int_1^{\infty} k e^{-2k^2}dk = \frac{1}{2}e^{-2} + \frac{1}{2}\int_1^{\infty} e^{-2k^2}dk.$$
Now we focus on the remaining improper integral. Because it is a Gaussian integral, we know it converges. Thus, the entire improper integral converges.
03
Determine the convergence or divergence of the series
Since the improper integral \(\int_1^{\infty} k e^{-2 k^{2}} dk\) converges, then by the Integral Test, the series \(\sum_{k=1}^{\infty} k e^{-2 k^{2}}\) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Improper Integral
When we talk about improper integrals, we refer to integrals with one or more infinite limits of integration, or integrals of functions with infinite discontinuities. They are a way to describe the area under a curve when the region extends indefinitely or the function has certain types of unbounded behavior.
To compute an improper integral, we typically take the limit of a definite integral as one of the bounds goes to infinity, or as we approach the point of discontinuity. For example, the integral \(\int_1^{\infty} k e^{-2k^2}dk\) in our exercise is improper because the upper limit is infinity. We need to evaluate this as the limit of \(\int_1^{a} k e^{-2k^2}dk\) as \(a\) approaches infinity.
It's crucial for the result to exist: if the limit is a finite number, we say the improper integral converges. If the limit doesn't exist or is infinite, the integral diverges. The behavior of improper integrals is particularly important in the study of infinite series, where they can determine if a series converges or diverges.
To compute an improper integral, we typically take the limit of a definite integral as one of the bounds goes to infinity, or as we approach the point of discontinuity. For example, the integral \(\int_1^{\infty} k e^{-2k^2}dk\) in our exercise is improper because the upper limit is infinity. We need to evaluate this as the limit of \(\int_1^{a} k e^{-2k^2}dk\) as \(a\) approaches infinity.
It's crucial for the result to exist: if the limit is a finite number, we say the improper integral converges. If the limit doesn't exist or is infinite, the integral diverges. The behavior of improper integrals is particularly important in the study of infinite series, where they can determine if a series converges or diverges.
Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a positive constant called the base, and \(x\) is the exponent. These functions are widely used because they describe a vast number of phenomena, such as population growth, radioactive decay, and compound interest.
One essential property of exponential functions is that they rise or fall at rates proportional to their current value. This is reflected in their graphs, which show a constant rate of growth or decay.
In our context, \(e^{-2 k^{2}}\) is an exponential function with base \(e\), the famous irrational number approximately equal to 2.71828. This base is particularly significant in calculus because of its connection to the rate of change and continuous growth. The negative exponent indicates that the function represents exponential decay, which means the function's value gets smaller as \(k\) increases. This decay is faster than polynomial decline, which is why the term \(ke^{-2k^2}\) approaches zero as \(k\) grows large, affecting both the evaluation of the improper integral and the series' convergence.
One essential property of exponential functions is that they rise or fall at rates proportional to their current value. This is reflected in their graphs, which show a constant rate of growth or decay.
In our context, \(e^{-2 k^{2}}\) is an exponential function with base \(e\), the famous irrational number approximately equal to 2.71828. This base is particularly significant in calculus because of its connection to the rate of change and continuous growth. The negative exponent indicates that the function represents exponential decay, which means the function's value gets smaller as \(k\) increases. This decay is faster than polynomial decline, which is why the term \(ke^{-2k^2}\) approaches zero as \(k\) grows large, affecting both the evaluation of the improper integral and the series' convergence.
Series Convergence
When discussing series convergence, we're looking at whether the sum of an infinite series of terms reaches a finite limit or not. If there is a finite sum, we say the series converges; if the sum is infinite or oscillates without approaching any limit, the series diverges.
Deciding whether an infinite series converges can be challenging, but there are several tests that we can use. In the provided exercise, we use the Integral Test to determine convergence. This test compares a series to an improper integral; if the integral of the function from some point to infinity is finite (converges), then the related series also converges, and vice versa.
The Integral Test is valid under specific conditions: the series' terms must come from a function that is continuous, positive, and decreasing. If these conditions are fulfilled, as they are in our exercise with the function \(f(k) = k e^{-2 k^{2}}\), and the improper integral converges, it confirms that the series converges as well. By working through the steps methodically and showing that the integral is finite, we are able to conclude that the series \(\sum_{k=1}^{\infty} k e^{-2 k^{2}}\) converges, too.
Deciding whether an infinite series converges can be challenging, but there are several tests that we can use. In the provided exercise, we use the Integral Test to determine convergence. This test compares a series to an improper integral; if the integral of the function from some point to infinity is finite (converges), then the related series also converges, and vice versa.
The Integral Test is valid under specific conditions: the series' terms must come from a function that is continuous, positive, and decreasing. If these conditions are fulfilled, as they are in our exercise with the function \(f(k) = k e^{-2 k^{2}}\), and the improper integral converges, it confirms that the series converges as well. By working through the steps methodically and showing that the integral is finite, we are able to conclude that the series \(\sum_{k=1}^{\infty} k e^{-2 k^{2}}\) converges, too.
