Problem 22
Question
Test these series for (a) absolute convergence, (b) conditional convergence. \(\sum(-1)^{k} \frac{k^{2}}{2^{k}}\).
Step-by-Step Solution
Verified Answer
The given series \(\sum(-1)^{k} \frac{k^{2}}{2^{k}}\) converges absolutely after applying the Ratio Test on the series \(\sum\frac{k^{2}}{2^{k}}\). Therefore, there's no need to test for conditional convergence.
1Step 1: Test for absolute convergence
To test for absolute convergence, we need to determine if the series \(\sum\left|(-1)^{k}\frac{k^{2}}{2^{k}}\right|\) converges.
First, take the absolute value of the terms in the original series:
\(\left|(-1)^{k}\frac{k^{2}}{2^{k}}\right| = \frac{k^{2}}{2^{k}}\)
Now, we have the series \(\sum\frac{k^{2}}{2^{k}}\). To determine its convergence, we can use the Ratio Test:
- Calculate the limit as \(k\) approaches infinity for the ratio of consecutive terms:
\(\lim_{k\to\infty}\frac{a_{k+1}}{a_{k}} = \lim_{k\to\infty}\frac{(k+1)^{2}/2^{k+1}}{k^{2}/2^{k}}\)
2Step 2: Apply the Ratio Test
Compute the limit:
\(\lim_{k\to\infty}\frac{(k+1)^{2}/2^{k+1}}{k^{2}/2^{k}} = \lim_{k\to\infty}\frac{(k+1)^{2}}{k^{2}}\cdot\frac{2^{k}}{2^{k+1}}\)
Simplify the limit:
\(\lim_{k\to\infty}\frac{(k+1)^{2}}{k^{2}}\cdot\frac{1}{2} = \frac{1}{2}\lim_{k\to\infty}\frac{k^{2}+2k+1}{k^{2}}\)
Divide numerator and denominator by \(k^{2}\):
\(\frac{1}{2}\lim_{k\to\infty}\frac{1+2/k+1/k^{2}}{1} = \frac{1}{2}(1+0+0) = \frac{1}{2}\)
The limit is less than 1, so by the Ratio Test, the series \(\sum\frac{k^{2}}{2^{k}}\) converges. Thus, the given series converges absolutely.
3Step 3: Conclusion
Since the series converged absolutely, there is no reason to test for conditional convergence. The given series \(\sum(-1)^{k} \frac{k^{2}}{2^{k}}\) converges absolutely.
Key Concepts
Absolute ConvergenceConditional ConvergenceRatio Test
Absolute Convergence
When we talk about absolute convergence, we mean that a series converges even when we take the absolute value of all its terms. This is a stronger form of convergence.
To test for absolute convergence, you take the series \( \sum a_k \) and consider if \( \sum |a_k| \) also converges.
If this absolute valued series converges, then the original series is absolutely convergent.In the exercise, we start by considering the series \( \sum(-1)^{k} \frac{k^{2}}{2^{k}} \). To test for absolute convergence, we take the absolute value of the terms:- \( \left| (-1)^{k} \frac{k^{2}}{2^{k}} \right| = \frac{k^{2}}{2^{k}} \)The next step involves checking if this new series \( \sum \frac{k^{2}}{2^{k}} \) converges.
If this series converges, then the original series is absolutely convergent. In this case, by applying the ratio test, it was shown that this series converges, leading to the conclusion of absolute convergence for the original series.
To test for absolute convergence, you take the series \( \sum a_k \) and consider if \( \sum |a_k| \) also converges.
If this absolute valued series converges, then the original series is absolutely convergent.In the exercise, we start by considering the series \( \sum(-1)^{k} \frac{k^{2}}{2^{k}} \). To test for absolute convergence, we take the absolute value of the terms:- \( \left| (-1)^{k} \frac{k^{2}}{2^{k}} \right| = \frac{k^{2}}{2^{k}} \)The next step involves checking if this new series \( \sum \frac{k^{2}}{2^{k}} \) converges.
If this series converges, then the original series is absolutely convergent. In this case, by applying the ratio test, it was shown that this series converges, leading to the conclusion of absolute convergence for the original series.
Conditional Convergence
Conditional convergence happens when a series converges, but it does not converge absolutely. In simpler terms, it means that the series \( \sum a_k \) converges, but when you take the absolute values \( \sum |a_k| \), it does not converge.
This type of convergence can often occur in alternating series.In our exercise, once we test for absolute convergence and find the series is indeed absolutely convergent, there's no need to examine conditional convergence. This is because absolute convergence implies regular convergence, always.
In contrast, if the absolute series does not converge, then checking for regular convergence of the original series could determine if it's conditionally convergent.The clue here is understanding that conditional convergence is only of interest if absolute convergence does not hold. In this exercise, finding absolute convergence rendered conditional convergence irrelevant.
This type of convergence can often occur in alternating series.In our exercise, once we test for absolute convergence and find the series is indeed absolutely convergent, there's no need to examine conditional convergence. This is because absolute convergence implies regular convergence, always.
In contrast, if the absolute series does not converge, then checking for regular convergence of the original series could determine if it's conditionally convergent.The clue here is understanding that conditional convergence is only of interest if absolute convergence does not hold. In this exercise, finding absolute convergence rendered conditional convergence irrelevant.
Ratio Test
The Ratio Test is a tool used to determine the convergence or divergence of a series. It's particularly useful for series that contain factorials, exponential terms, or powers, making the test perfect for our exercise.To apply the Ratio Test:- Calculate the limit of the absolute value of the ratio of consecutive terms, \( \frac{a_{k+1}}{a_k} \).- Consider the limit as \( k \to \infty \).For example, in the exercise for the series \( \sum \frac{k^{2}}{2^{k}} \), we find this limit:- First, form the ratio \( \frac{(k+1)^2/2^{k+1}}{k^2/2^k} \).- Simplify it to \( \frac{(k+1)^2}{k^2} \cdot \frac{1}{2} \).- Then find \( \lim_{k \to \infty} \frac{(k^2 + 2k + 1)}{k^2} \cdot \frac{1}{2} = \frac{1}{2} \).Since the result is less than 1, the series \( \sum \frac{k^{2}}{2^{k}} \) converges by the Ratio Test. When applying this test, a limit less than 1 guarantees convergence, a limit more than 1 indicates divergence, and a limit equal to 1 means the test is inconclusive.
Other exercises in this chapter
Problem 22
Expand \(g(x)\) as indicated and specify the values of \(x\) for which the expansion is valid. \(g(x)=\ln (2+3 x)\) in powers of \(x-4\).
View solution Problem 22
Determine whether the series converges or diverges. $$\sum \frac{(k !)^{2}}{(2 k) !}$$
View solution Problem 22
Determine whether the series converges or diverse. $$\sum \frac{1}{2^{k+1}-1}$$
View solution Problem 22
Find the interval of convergence. $$\sum k ! x^{k}$$
View solution