`lim_(n -> oo) (((n+1)(n+2)(n+3).......2n) / n^(2n))^(1/n)`is equal to

Options

A. `27/(e^(2)0`
B. `9/(e^(2))`
C. `3log3-2`
D. `18/e^(4)`

Explanation

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{(n+1)(n+2)(n+3)\cdots(2n)}{n^{2n}} \right)^{\frac{1}{n}}, \]

we can follow these steps:

### Step 1: Rewrite the limit
Define
\[ L = \lim_{n \to \infty} \left( \frac{(n+1)(n+2)(n+3)\cdots(2n)}{n^{2n}} \right)^{\frac{1}{n}}. \]
We can express the numerator as a product of terms from \(n+1\) to \(2n\).

### Step 2: Express the product in terms of factorials
The product \((n+1)(n+2)(n+3)\cdots(2n)\) can be rewritten using factorials:
\[ (n+1)(n+2)(n+3)\cdots(2n) = \frac{(2n)!}{n!}. \]
Thus, we can rewrite \(L\) as:
\[ L = \lim_{n \to \infty} \left( \frac{(2n)!}{n! \cdot n^{2n}} \right)^{\frac{1}{n}}. \]

### Step 3: Take the logarithm
To simplify the limit, we take the natural logarithm:
\[ \log L = \lim_{n \to \infty} \frac{1}{n} \left( \log(2n)! - \log(n!) - 2n \log n \right). \]

### Step 4: Use Stirling's approximation
Using Stirling's approximation, \(\log k! \approx k \log k - k\), we can approximate:
\[ \log(2n)! \approx 2n \log(2n) - 2n = 2n \log 2 + 2n \log n - 2n, \]
and
\[ \log(n!) \approx n \log n - n. \]

### Step 5: Substitute the approximations
Substituting these approximations into our expression for \(\log L\):
\[ \log L \approx \lim_{n \to \infty} \frac{1}{n} \left( (2n \log 2 + 2n \log n - 2n) - (n \log n - n) - 2n \log n \right). \]
Simplifying this gives:
\[ \log L \approx \lim_{n \to \infty} \frac{1}{n} \left( 2n \log 2 + 2n \log n - 2n - n \log n + n - 2n \log n \right). \]
This simplifies to:
\[ \log L \approx \lim_{n \to \infty} \frac{1}{n} \left( 2n \log 2 - n \right) = 2 \log 2 - 1. \]

### Step 6: Exponentiate to find \(L\)
Now, exponentiating both sides gives:
\[ L = e^{2 \log 2 - 1} = \frac{4}{e}. \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to \infty} \left( \frac{(n+1)(n+2)(n+3)\cdots(2n)}{n^{2n}} \right)^{\frac{1}{n}} = \frac{4}{e}. \]