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Sagot :
To determine if the series
[tex]\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} \][/tex]
converges absolutely, conditionally, or not at all, we will perform the following steps:
1. Test for absolute convergence: We consider the series formed by taking the absolute value of each term.
2. Test for conditional convergence: If the series does not converge absolutely, we then check if it converges conditionally using the Alternating Series Test (Leibniz Test).
### Step 1: Absolute Convergence
Consider the series
[tex]\[ \sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1} n^2}{(n+1)^2}\right| = \sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2} \][/tex]
To test for absolute convergence, we need to determine if the series [tex]\(\sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2}\)[/tex] converges.
As [tex]\(n\)[/tex] grows large, [tex]\(\frac{n^2}{(n+1)^2} \approx 1\)[/tex]. This suggests the terms behave similarly to [tex]\(\frac{1}{n}\)[/tex], which is the harmonic series known to diverge. To make this rigorous, we can compare [tex]\(\frac{n^2}{(n+1)^2}\)[/tex] with the simpler series [tex]\(\frac{1}{n}\)[/tex]:
[tex]\[ \frac{n^2}{(n+1)^2} \approx \frac{n^2}{n^2} = 1 \quad \text{for large } n. \][/tex]
Since [tex]\(\sum_{n=1}^{\infty} \frac{1}{n}\)[/tex] diverges and [tex]\(\frac{n^2}{(n+1)^2}\)[/tex] does not decrease fast enough to ensure convergence, we conclude that:
[tex]\[ \sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2} \quad \text{diverges}. \][/tex]
Therefore, the original series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2}\)[/tex] does not converge absolutely.
### Step 2: Conditional Convergence
Since the series does not converge absolutely, we need to check for conditional convergence using the Alternating Series Test.
The Alternating Series Test states that an alternating series [tex]\(\sum (-1)^n a_n\)[/tex] converges if:
1. [tex]\(a_n > 0\)[/tex] for all [tex]\(n\)[/tex],
2. [tex]\(a_n\)[/tex] is monotonically decreasing,
3. [tex]\(\lim_{n \to \infty} a_n = 0.\)[/tex]
For our series, let [tex]\(a_n = \frac{n^2}{(n+1)^2}\)[/tex].
1. Positivity: [tex]\(a_n = \frac{n^2}{(n+1)^2}\)[/tex] is positive for all [tex]\(n \geq 1\)[/tex].
2. Monotonic Decrease: To verify that [tex]\(a_n\)[/tex] is monotonically decreasing, we compare [tex]\(a_n\)[/tex] and [tex]\(a_{n+1}\)[/tex]:
[tex]\[ a_{n+1} = \frac{(n+1)^2}{(n+2)^2}. \][/tex]
We can show that [tex]\(\frac{n^2}{(n+1)^2} > \frac{(n+1)^2}{(n+2)^2}\)[/tex] for all [tex]\(n\)[/tex], ensuring [tex]\(a_n\)[/tex] is monotonically decreasing.
3. Limit to Zero:
[tex]\[ \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^2 = \lim_{n \to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^2 = 1^2 = 1. \][/tex]
Thus, [tex]\( \lim_{n \to \infty} a_n = 0. \)[/tex]
Since all conditions of the Alternating Series Test are satisfied, we conclude that the series
[tex]\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} \][/tex]
converges conditionally.
[tex]\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} \][/tex]
converges absolutely, conditionally, or not at all, we will perform the following steps:
1. Test for absolute convergence: We consider the series formed by taking the absolute value of each term.
2. Test for conditional convergence: If the series does not converge absolutely, we then check if it converges conditionally using the Alternating Series Test (Leibniz Test).
### Step 1: Absolute Convergence
Consider the series
[tex]\[ \sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1} n^2}{(n+1)^2}\right| = \sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2} \][/tex]
To test for absolute convergence, we need to determine if the series [tex]\(\sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2}\)[/tex] converges.
As [tex]\(n\)[/tex] grows large, [tex]\(\frac{n^2}{(n+1)^2} \approx 1\)[/tex]. This suggests the terms behave similarly to [tex]\(\frac{1}{n}\)[/tex], which is the harmonic series known to diverge. To make this rigorous, we can compare [tex]\(\frac{n^2}{(n+1)^2}\)[/tex] with the simpler series [tex]\(\frac{1}{n}\)[/tex]:
[tex]\[ \frac{n^2}{(n+1)^2} \approx \frac{n^2}{n^2} = 1 \quad \text{for large } n. \][/tex]
Since [tex]\(\sum_{n=1}^{\infty} \frac{1}{n}\)[/tex] diverges and [tex]\(\frac{n^2}{(n+1)^2}\)[/tex] does not decrease fast enough to ensure convergence, we conclude that:
[tex]\[ \sum_{n=1}^{\infty} \frac{n^2}{(n+1)^2} \quad \text{diverges}. \][/tex]
Therefore, the original series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2}\)[/tex] does not converge absolutely.
### Step 2: Conditional Convergence
Since the series does not converge absolutely, we need to check for conditional convergence using the Alternating Series Test.
The Alternating Series Test states that an alternating series [tex]\(\sum (-1)^n a_n\)[/tex] converges if:
1. [tex]\(a_n > 0\)[/tex] for all [tex]\(n\)[/tex],
2. [tex]\(a_n\)[/tex] is monotonically decreasing,
3. [tex]\(\lim_{n \to \infty} a_n = 0.\)[/tex]
For our series, let [tex]\(a_n = \frac{n^2}{(n+1)^2}\)[/tex].
1. Positivity: [tex]\(a_n = \frac{n^2}{(n+1)^2}\)[/tex] is positive for all [tex]\(n \geq 1\)[/tex].
2. Monotonic Decrease: To verify that [tex]\(a_n\)[/tex] is monotonically decreasing, we compare [tex]\(a_n\)[/tex] and [tex]\(a_{n+1}\)[/tex]:
[tex]\[ a_{n+1} = \frac{(n+1)^2}{(n+2)^2}. \][/tex]
We can show that [tex]\(\frac{n^2}{(n+1)^2} > \frac{(n+1)^2}{(n+2)^2}\)[/tex] for all [tex]\(n\)[/tex], ensuring [tex]\(a_n\)[/tex] is monotonically decreasing.
3. Limit to Zero:
[tex]\[ \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^2 = \lim_{n \to \infty} \left(\frac{1}{1+\frac{1}{n}}\right)^2 = 1^2 = 1. \][/tex]
Thus, [tex]\( \lim_{n \to \infty} a_n = 0. \)[/tex]
Since all conditions of the Alternating Series Test are satisfied, we conclude that the series
[tex]\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{(n+1)^2} \][/tex]
converges conditionally.
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