Problem 21
Question
Test these series for (a) absolute convergence, (b) conditional convergence. \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\).
Step-by-Step Solution
Verified Answer
The given series is \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\). We find the absolute value of the series to be \(\frac{4^{k-2}}{e^{k}}\). By performing the Ratio Test, we find the limit \(\lim_{k \to \infty}\frac{4}{e}\), which is less than 1. Therefore, the series converges absolutely. Since the series converges absolutely, it also converges conditionally.
1Step 1: Find the absolute value of the series
First, let's find the absolute value of the series:
\(|(-1)^{k} \frac{4^{k-2}}{e^{k}}| = \frac{4^{k-2}}{e^{k}}\)
2Step 2: Test for absolute convergence
To test absolute convergence, we'll test if the absolute series, \(\sum \frac{4^{k-2}}{e^{k}}\), converges. We can perform the Ratio Test, which states that if \(\lim_{k \to \infty} \frac{a_{k+1}}{a_{k}} < 1\), then the series converges absolutely.
Calculate the limit:
\(\lim_{k \to \infty} \frac{\frac{4^{(k+1)-2}}{e^{k+1}}}{\frac{4^{k-2}}{e^{k}}}\)
3Step 3: Simplify the limit
Simplify the limit by canceling common terms:
\(\lim_{k \to \infty}\frac{4^{k-1}e^k}{4^{k-2}e^{k+1}} = \lim_{k \to \infty}\frac{4}{e}\)
Since the limit is \(\frac{4}{e}\) which is less than 1, the given series converges absolutely.
4Step 4: Test for conditional convergence
Since the series converges absolutely, it also converges conditionally. Therefore, we have determined that the series converges both absolutely and conditionally.
Key Concepts
Absolute ConvergenceConditional ConvergenceRatio Test
Absolute Convergence
Understanding absolute convergence is pivotal when dealing with infinite series. In our case, the exercise requires analyzing a series to determine if it is absolutely convergent.
Absolute convergence of a series occurs when the sum of the absolute values of its terms converges. Mathematically, for a given series \(\sum a_k\), if the series \(\sum |a_k|\) converges, then the original series \(\sum a_k\) is said to be absolutely convergent.
Now, why does this matter? An absolutely convergent series is robust in the sense that altering the order of its terms does not affect its sum. This is not necessarily true for series that are not absolutely convergent. In the case of the exercise, the series \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\) was tested by first considering the absolute series \(\sum \frac{4^{k-2}}{e^{k}}\). By employing the Ratio Test and finding that the limit yields a value less than 1, we confirm that the series is absolutely convergent.
Absolute convergence of a series occurs when the sum of the absolute values of its terms converges. Mathematically, for a given series \(\sum a_k\), if the series \(\sum |a_k|\) converges, then the original series \(\sum a_k\) is said to be absolutely convergent.
Now, why does this matter? An absolutely convergent series is robust in the sense that altering the order of its terms does not affect its sum. This is not necessarily true for series that are not absolutely convergent. In the case of the exercise, the series \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\) was tested by first considering the absolute series \(\sum \frac{4^{k-2}}{e^{k}}\). By employing the Ratio Test and finding that the limit yields a value less than 1, we confirm that the series is absolutely convergent.
Conditional Convergence
The concept of conditional convergence arises when considering series that converge but do not meet the criteria for absolute convergence. In simpler terms, a conditionally convergent series is one where \(\sum a_k\) converges, but \(\sum |a_k|\) does not.
This distinction is crucial because conditionally convergent series can lead to different sums when their terms are rearranged - a property not seen in absolutely convergent series. It's a phenomenon famously illustrated by the Riemann series theorem.
In our original exercise, the series \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\) was subject to the test for absolute convergence. Since it was found to be absolutely convergent, it is also conditionally convergent by definition. This is because conditional convergence is a broader classification: if a series is absolutely convergent, then it is also convergently, but not every conditionally convergent series is absolutely convergent.
This distinction is crucial because conditionally convergent series can lead to different sums when their terms are rearranged - a property not seen in absolutely convergent series. It's a phenomenon famously illustrated by the Riemann series theorem.
In our original exercise, the series \(\sum(-1)^{k} \frac{4^{k-2}}{e^{k}}\) was subject to the test for absolute convergence. Since it was found to be absolutely convergent, it is also conditionally convergent by definition. This is because conditional convergence is a broader classification: if a series is absolutely convergent, then it is also convergently, but not every conditionally convergent series is absolutely convergent.
Ratio Test
The Ratio Test is a powerful tool for determining the convergence of series, especially when the series' terms involve exponential functions, factorials, or powers.
The Ratio Test states that for a series \(\sum a_k\), we compute the limit \(L = \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_{k}|}\). If \(L < 1\), the series converges absolutely; if \(L > 1\) or the limit is infinite, the series diverges; and if \(L = 1\), the test is inconclusive.
In our exercise, we applied the Ratio Test to \(\sum |(-1)^{k} \frac{4^{k-2}}{e^{k}}|\) which converted to \(\sum \frac{4^{k-2}}{e^{k}}\), leading to the limit \(\lim_{k \to \infty} \frac{4}{e}\). Since this limit is clearly less than 1, we conclude that the series converges absolutely. The Ratio Test is indeed an effective technique for students to master as it provides a clear-cut way to assess the behavior of series.
The Ratio Test states that for a series \(\sum a_k\), we compute the limit \(L = \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_{k}|}\). If \(L < 1\), the series converges absolutely; if \(L > 1\) or the limit is infinite, the series diverges; and if \(L = 1\), the test is inconclusive.
In our exercise, we applied the Ratio Test to \(\sum |(-1)^{k} \frac{4^{k-2}}{e^{k}}|\) which converted to \(\sum \frac{4^{k-2}}{e^{k}}\), leading to the limit \(\lim_{k \to \infty} \frac{4}{e}\). Since this limit is clearly less than 1, we conclude that the series converges absolutely. The Ratio Test is indeed an effective technique for students to master as it provides a clear-cut way to assess the behavior of series.
Other exercises in this chapter
Problem 21
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View solution