To determine the convergence of the given series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\)[/tex], we can use the Alternating Series Test (Leibniz's Test) and additional tests for absolute convergence.
### Step 1: Identify the [tex]\(b_n\)[/tex] and check the conditions for the Alternating Series Test
The series is given by:
[tex]\[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \][/tex]
Here, the term [tex]\(a_n = \frac{(-1)^n}{n}\)[/tex] and thus [tex]\(b_n = \frac{1}{n}\)[/tex].
The Alternating Series Test states that an alternating series [tex]\(\sum (-1)^n b_n\)[/tex] converges if:
1. [tex]\( b_n \)[/tex] is positive.
2. [tex]\( b_n \)[/tex] is monotonically decreasing, i.e., [tex]\( b_{n+1} \le b_n \)[/tex].
3. [tex]\(\lim_{n \to \infty} b_n = 0\)[/tex].
### Step 2: Check the Limit
Evaluate the following limit:
[tex]\[ \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{n} \][/tex]
As [tex]\(n\)[/tex] approaches infinity, [tex]\(\frac{1}{n}\)[/tex] approaches 0.
[tex]\[ \lim_{n \to \infty} \frac{1}{n} = 0 \][/tex]
### Step 3: Check if [tex]\(b_n\)[/tex] is Positive and Monotonically Decreasing
1. [tex]\(b_n = \frac{1}{n}\)[/tex] is positive for all [tex]\(n \ge 1\)[/tex].
2. [tex]\(b_n = \frac{1}{n}\)[/tex] is monotonically decreasing because as [tex]\(n\)[/tex] increases, [tex]\(\frac{1}{n}\)[/tex] decreases.
Since [tex]\(b_n\)[/tex] is positive, decreasing, and its limit as [tex]\(n\)[/tex] approaches infinity is 0, we can conclude that the series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\)[/tex] converges by the Alternating Series Test.
### Step 4: Check for Absolute Convergence
To determine if the series is absolutely convergent, consider the absolute value series:
[tex]\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n} \right| = \sum_{n=1}^{\infty} \frac{1}{n} \][/tex]
The series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n}\)[/tex] is the harmonic series, which is known to diverge.
### Step 5: Conclusion on Absolute and Conditional Convergence
Since the absolute value series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n}\)[/tex] diverges, the original series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\)[/tex] is not absolutely convergent. However, it does converge conditionally by the Alternating Series Test.
### Final Conclusion
By the Alternating Series Test, the series [tex]\(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\)[/tex] converges conditionally.
Therefore, the series is:
- Conditionally convergent.
