Let `alpha(n)=1-(1)/(2)+(1)/(3)-(1)/(4)+.........+(-1)^(n-1)(1)/(n),ninN`, then

Options

A. `(1)/(n+1)+(1)/(n+2)+.........+(1)/(2n)=alpha(2n)`
B. `alpha(2n)lt1AAninN`
C. `alpha(2n)ge(1)/(2)AAninN`
D. `alpha(n)gt(1)/(2)AAninN`

Answer

To solve the problem, we need to analyze the series defined by \( \alpha(n) \):

\[ \alpha(n) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + (-1)^{n-1} \frac{1}{n} \]

This series is an alternating series, and we want to evaluate \( \alpha(n) \) for various values of \( n \) and determine its properties.

### Step 1: Understanding the Series
The series can be rewritten as:

\[ \alpha(n) = \sum_{k=1}^{n} (-1)^{k-1} \frac{1}{k} \]

This means we are summing the terms \( \frac{1}{k} \) with alternating signs.

### Step 2: Evaluating \( \alpha(2n) \)
Let's first evaluate \( \alpha(2n) \):

\[ \alpha(2n) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + (-1)^{2n-1} \frac{1}{2n} \]

### Step 3: Grouping the Terms
We can group the terms in pairs:

\[ \alpha(2n) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n}\right) \]

Each pair \( \left(\frac{1}{k} - \frac{1}{k+1}\right) \) is positive and contributes to the sum.

### Step 4: Analyzing the Limit
As \( n \) increases, the series converges to \( \ln(2) \). Therefore, we can say:

\[ \alpha(2n) \to \ln(2) \text{ as } n \to \infty \]

### Step 5: Establishing Bounds
We know that \( \ln(2) \approx 0.693 \), which is less than 1. Hence, we can conclude that:

\[ \alpha(2n) < 1 \]

### Step 6: Evaluating \( \alpha(n) \)
Now we evaluate \( \alpha(n) \):

- For odd \( n \), the last term is positive, and for even \( n \), the last term is negative.
- Therefore, \( \alpha(n) \) oscillates but remains bounded between 0 and 1.

### Conclusion
From our analysis, we can summarize:

- \( \alpha(2n) < 1 \)
- \( \alpha(n) \) is bounded and tends towards \( \ln(2) \), which is greater than 0.5.

Thus, we can conclude:

\[ \alpha(n) > \frac{1}{2} \]

### Final Result
The properties of \( \alpha(n) \) lead us to the conclusion that:

- \( \alpha(2n) < 1 \)
- \( \alpha(n) > \frac{1}{2} \)