Let `a_(n) = (1+1/n)^(n) .` Then for each `n in N`

Options

A. `a_(n)ge 2`
B. `a_(n)lt 3`
C. `a_(n)lt 4`
D. `a_(n)lt 2`

Explanation

To solve the problem, we need to analyze the expression \( a_n = \left(1 + \frac{1}{n}\right)^n \) and determine its behavior as \( n \) varies over the natural numbers \( \mathbb{N} \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression \( a_n = \left(1 + \frac{1}{n}\right)^n \) is a well-known limit that approaches the mathematical constant \( e \) as \( n \) approaches infinity.

2. **Using the Binomial Theorem**:
We can expand \( a_n \) using the Binomial Theorem:
\[ a_n = \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{n}\right)^k \]
This expansion gives us:
\[ a_n = \binom{n}{0} + \binom{n}{1} \cdot \frac{1}{n} + \binom{n}{2} \cdot \left(\frac{1}{n}\right)^2 + \ldots + \binom{n}{n} \cdot \left(\frac{1}{n}\right)^n \]

3. **Calculating the First Few Terms**:
The first few terms of the expansion are:
- \( \binom{n}{0} = 1 \)
- \( \binom{n}{1} \cdot \frac{1}{n} = 1 \)
- Higher-order terms become smaller as \( n \) increases.

Thus, we can approximate:
\[ a_n \approx 1 + 1 + \text{(smaller terms)} \]

4. **Taking the Limit as \( n \to \infty \)**:
As \( n \) approaches infinity, the expression converges to:
\[ \lim_{n \to \infty} a_n = e \approx 2.718 \]

5. **Establishing Bounds**:
We can establish that:
\[ 2 < a_n < e \quad \text{for all } n \in \mathbb{N} \]
This means \( a_n \) is always greater than 2 and less than \( e \).

6. **Conclusion**:
Therefore, we can conclude that:
- \( a_n > 2 \)
- \( a_n < 3 \) (since \( e < 3 \))
- \( a_n < 4 \)

### Final Answer:
The correct options based on the analysis are:
- \( a_n > 2 \)
- \( a_n < 3 \)
- \( a_n < 4 \)