What are the limits of `(2^(n)(n+1)^(n))/(n^(n))`, where n is a positive integer ?

Options

A. `2 le (2^(n)(n+1)^(n))/(n^(n)) le 3`
B. `2^(n+1) le (2^(n)(n+1)^(n))/(n^(n)) le 3.2^(n)`
C. `2^(n) le (2^(n)(n+1)^(n))/(n^(n)) le 3^(n)`
D. None of these

Answer

To find the limits of the expression \(\frac{2^n (n+1)^n}{n^n}\) where \(n\) is a positive integer, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ f(n) = \frac{2^n (n+1)^n}{n^n} \]
This can be rewritten as:
\[ f(n) = 2^n \cdot \left(\frac{n+1}{n}\right)^n \]

### Step 2: Simplify the fraction
The fraction \(\frac{n+1}{n}\) simplifies to:
\[ \frac{n+1}{n} = 1 + \frac{1}{n} \]
Thus, we can express \(f(n)\) as:
\[ f(n) = 2^n \cdot \left(1 + \frac{1}{n}\right)^n \]

### Step 3: Analyze the limit of \(\left(1 + \frac{1}{n}\right)^n\)
As \(n\) approaches infinity, we know from calculus that:
\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \]
Therefore, we can say:
\[ \left(1 + \frac{1}{n}\right)^n \to e \text{ as } n \to \infty \]

### Step 4: Combine the limits
Now substituting back into our expression for \(f(n)\):
\[ f(n) \approx 2^n \cdot e \text{ as } n \to \infty \]

### Step 5: Analyze the behavior as \(n\) increases
As \(n\) increases, \(2^n\) grows exponentially. Therefore, the limit of \(f(n)\) as \(n\) approaches infinity is:
\[ \lim_{n \to \infty} f(n) = \infty \]

### Step 6: Check the value at \(n = 1\)
To find the minimum limit, we can check the value of \(f(n)\) at \(n = 1\):
\[ f(1) = \frac{2^1 (1+1)^1}{1^1} = \frac{2 \cdot 2}{1} = 4 \]

### Conclusion
Thus, the limits of the expression \(\frac{2^n (n+1)^n}{n^n}\) where \(n\) is a positive integer are:
- The minimum value is \(4\) when \(n = 1\).
- The limit approaches infinity as \(n\) approaches infinity.