The value of `lim_(n rarr oo) [((2n)!)/(n!n^(n))]^(1//n)` is equal to

Options

A. 4e
B. e/4
C. 4/e
D. None of these

Answer

To solve the limit \( \lim_{n \to \infty} \left( \frac{(2n)!}{n! n^n} \right)^{\frac{1}{n}} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the expression:
\[ L = \lim_{n \to \infty} \left( \frac{(2n)!}{n! n^n} \right)^{\frac{1}{n}} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \log \left( \frac{(2n)!}{n! n^n} \right) \]
This can be simplified to:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \log(2n)! - \log(n!) - n \log(n) \right) \]

### Step 3: Use Stirling's approximation
Using Stirling's approximation, \( \log(k!) \approx k \log(k) - k \), we can approximate the logarithms:
\[ \log(2n)! \approx 2n \log(2n) - 2n = 2n (\log(2) + \log(n)) - 2n = 2n \log(2) + 2n \log(n) - 2n \]
\[ \log(n!) \approx n \log(n) - n \]

### Step 4: Substitute the approximations
Substituting these approximations into our expression for \( \log L \):
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \left( 2n \log(2) + 2n \log(n) - 2n \right) - \left( n \log(n) - n \right) - n \log(n) \right) \]

### Step 5: Simplify the expression
This simplifies to:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( 2n \log(2) + 2n \log(n) - 2n - n \log(n) + n - n \log(n) \right) \]
\[ = \lim_{n \to \infty} \frac{1}{n} \left( 2n \log(2) + n \log(n) - n \right) \]
\[ = \lim_{n \to \infty} \left( 2 \log(2) + \log(n) - 1 \right) \]

### Step 6: Evaluate the limit
As \( n \to \infty \), \( \log(n) \to \infty \), thus:
\[ \log L \to \infty \]
This implies that \( L \to e^{\infty} = \infty \).

### Step 7: Final result
Thus, the limit evaluates to:
\[ \lim_{n \to \infty} \left( \frac{(2n)!}{n! n^n} \right)^{\frac{1}{n}} = 4 \]

### Conclusion
The final answer is:
\[ \boxed{4} \]