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Magazine Chapter 9: Problem 25
Use the Integral Test to determine the convergence or divergence of the following series, or state that the conditions of the test are not satisfied and, therefore, the test does not apply. $$\sum_{k=1}^{\infty} \frac{k}{e^{k}}$$
Short Answer
Expert verified
Answer: The infinite series converges.
Step by step solution
01
Check the function's conditions
We must first verify that the given function, f(k) = \(\frac{k}{e^k}\), is positive, continuous, and decreasing for k 鈮� 1.
1. Positive: Since k is a positive integer and e^k is a positive exponential function, the function f(k) must be positive for all k 鈮� 1.
2. Continuous: The function f(k) is a ratio of polynomials where the denominator is nonzero (since e^k > 0), so it is continuous for all k 鈮� 1.
3. Decreasing: We can examine the derivative of the function f(k) to see if it's decreasing for k 鈮� 1.
02
Calculate the derivative
To determine whether f(k) is decreasing for k 鈮� 1, we must calculate the derivative of the function:
$$f(k) = \frac{k}{e^k}$$
Using the quotient rule, we get:
$$f'(k) = \frac{e^k(-k) - (1)(k e^{k})}{(e^k)^{2}} = \frac{-2k e^k}{e^{2k}}$$
03
Analyze the function's behavior
Now that we have the derivative, we can analyze the behavior of the function:
$$f'(k) = \frac{-2k e^k}{e^{2k}}$$
Since -2k and e^{2k} are both positive for k 鈮� 1, the derivative f'(k) will be negative for all k 鈮� 1, showing that the function f(k) is decreasing for k 鈮� 1.
Now that we have confirmed that f(k) is positive, continuous, and decreasing for k 鈮� 1, we can proceed with the Integral Test.
04
Calculate the improper integral
Apply the Integral Test by calculating the improper integral:
$$\int_{1}^{\infty} \frac{k}{e^k} dk$$
Let's use the integration by parts method. We set u = k and dv = \(\frac{1}{e^k} dk\). Then, we compute du and v:
$$du = dk$$
$$v = -e^{-k}$$
Now, we can apply the integration by parts formula:
$$\int_{1}^{\infty} \frac{k}{e^k} dk = \left[-ke^{-k}\right]_{1}^{\infty} + \int_{1}^{\infty} e^{-k} dk$$
05
Evaluate the limit of the improper integral and determine convergence
We must now evaluate the limit of the improper integral:
$$\lim_{b\to\infty}\left[\left(-be^{-b}+\frac{e^{-b}}{e}\right)-\left(-e^{-1}+\frac{e^{-1}}{e}\right)\right]$$
Simplify the above expression to determine whether the limit exists or not:
$$\lim_{b\to\infty} \left(-\frac{b}{e^b}\right) = \lim_{b\to\infty} \frac{-b}{e^b} = 0$$
The first term approaches 0 as b goes to infinity because the exponential function e^b grows much faster than the linear function b. Therefore:
$$\lim_{b\to\infty}\left[\left(-be^{-b}+\frac{e^{-b}}{e}\right)-\left(-e^{-1}+\frac{e^{-1}}{e}\right)\right] = -1 + \frac{1}{e} < \infty$$
Since the limit is finite, the improper integral converges. According to the Integral Test, if the improper integral converges, then the given series converges as well.
06
Conclusion
By using the Integral Test, we have determined that the given series:
$$\sum_{k=1}^{\infty} \frac{k}{e^{k}}$$
converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence and Divergence
In the realm of infinite series, understanding whether a series converges or diverges is crucial. The term **convergence** refers to an infinite series whose sum approaches a finite value as the number of terms increases. On the other hand, **divergence** indicates that the series does not approach a finite limit, potentially growing indefinitely or oscillating without settling.
For students tackling the process of determining a series' behavior, the Integral Test is an invaluable tool. By transforming the series expression into a corresponding continuous function, this test helps determine if the integral of this function converges. If the integral converges, so does the series; if it diverges, likewise does the series.
For students tackling the process of determining a series' behavior, the Integral Test is an invaluable tool. By transforming the series expression into a corresponding continuous function, this test helps determine if the integral of this function converges. If the integral converges, so does the series; if it diverges, likewise does the series.
- **If the integral from 1 to infinity of the function converges**: The series converges.
- **If the integral from 1 to infinity of the function diverges**: The series diverges.
Improper Integrals
Improper integrals are a fascinating concept in calculus, used specifically when dealing with infinite limits of integration or unbounded functions. They help us analyze cases where normal integration techniques fall short. For the exercise at hand, we dealt with an improper integral of the form \(\int_{1}^{\infty} \frac{k}{e^k} \ dk\). This integral is termed 'improper' because of its infinite upper limit.
One prominent method of evaluating improper integrals is by taking the limit of a definite integral:
When applying the Integral Test to determine series convergence, the outcome of the improper integral directly indicates the behavior of the original series. In this example, the integral converged, confirming that the series \(\sum_{k=1}^{\infty} \frac{k}{e^{k}}\) converges as well.
One prominent method of evaluating improper integrals is by taking the limit of a definite integral:
- Replacing infinity with a variable \(b\), integrate from 1 to \(b\).
- After integration, compute the limit as \(b\) approaches infinity.
When applying the Integral Test to determine series convergence, the outcome of the improper integral directly indicates the behavior of the original series. In this example, the integral converged, confirming that the series \(\sum_{k=1}^{\infty} \frac{k}{e^{k}}\) converges as well.
Decreasing Functions
When applying the Integral Test, one must verify that the function representing the series is decreasing. A function is generally considered decreasing if each term is less than the preceding term over its interval of interest.
To verify a function's decreasing nature, examine its derivative. For the function \(f(k) = \frac{k}{e^k}\), the derivative \(f'(k) = \frac{-2k e^k}{e^{2k}}\) was calculated using the quotient rule. This derivative is negative for \(k \geq 1\), indicating that \(f(k)\) is decreasing because the negative derivative signifies a decline in value as \(k\) increases.
Consistently, these checks affirm a functional foundation for series analysis using the Integral Test. Ensuring that the function is positive, continuous, and decreasing sets the stage for reliable convergence assessments. This triple verification of conditions before proceeding with the Integral Test ensures that our subsequent conclusions are both valid and accurate.
To verify a function's decreasing nature, examine its derivative. For the function \(f(k) = \frac{k}{e^k}\), the derivative \(f'(k) = \frac{-2k e^k}{e^{2k}}\) was calculated using the quotient rule. This derivative is negative for \(k \geq 1\), indicating that \(f(k)\) is decreasing because the negative derivative signifies a decline in value as \(k\) increases.
Consistently, these checks affirm a functional foundation for series analysis using the Integral Test. Ensuring that the function is positive, continuous, and decreasing sets the stage for reliable convergence assessments. This triple verification of conditions before proceeding with the Integral Test ensures that our subsequent conclusions are both valid and accurate.
