The value of `lim_(n rarr infty) (1)/(n) {(n+)(n+2)(n+3)…(n+n)}^(1//n)` is equal to

Options

A. `4e`
B. `(e)/(4)`
C. `(4)/(e)`
D. None of these

Answer

To find the value of

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

we will follow these steps:

### Step 1: Rewrite the product
The expression inside the limit can be rewritten as:

\[ (n+1)(n+2)(n+3) \cdots (n+n) = \prod_{r=1}^{n} (n+r). \]

Thus, we can express our limit as:

\[ \lim_{n \to \infty} \frac{1}{n} \left( \prod_{r=1}^{n} (n+r) \right)^{\frac{1}{n}}. \]

### Step 2: Simplify the product
Now, we can factor out \(n\) from each term in the product:

\[ \prod_{r=1}^{n} (n+r) = \prod_{r=1}^{n} n(1 + \frac{r}{n}) = n^n \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right). \]

### Step 3: Substitute back into the limit
Substituting this back into our limit gives:

\[ \lim_{n \to \infty} \frac{1}{n} \left( n^n \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{n} \left( n^n \right)^{\frac{1}{n}} \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}}. \]

This simplifies to:

\[ \lim_{n \to \infty} \frac{1}{n} \cdot n \cdot \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}} = \lim_{n \to \infty} \left( \prod_{r=1}^{n} \left(1 + \frac{r}{n}\right) \right)^{\frac{1}{n}}. \]

### Step 4: Convert product to sum
Now, we can take the logarithm of the product:

\[ \log y = \frac{1}{n} \sum_{r=1}^{n} \log\left(1 + \frac{r}{n}\right). \]

As \(n \to \infty\), this sum approaches the integral:

\[ \log y \approx \int_{0}^{1} \log(1+x) \, dx. \]

### Step 5: Evaluate the integral
To evaluate the integral:

\[ \int \log(1+x) \, dx, \]

we can use integration by parts. Let \(u = \log(1+x)\) and \(dv = dx\). Then \(du = \frac{1}{1+x} dx\) and \(v = x\). Thus,

\[ \int \log(1+x) \, dx = x \log(1+x) - \int \frac{x}{1+x} \, dx. \]

The second integral can be simplified:

\[ \int \frac{x}{1+x} \, dx = \int \left(1 - \frac{1}{1+x}\right) dx = x - \log(1+x). \]

Putting it all together, we have:

\[ \int \log(1+x) \, dx = x \log(1+x) - (x - \log(1+x)) = (x + 1) \log(1+x) - x. \]

Now, we evaluate from 0 to 1:

\[ \left[(x + 1) \log(1+x) - x\right]_{0}^{1} = \left[2 \log(2) - 1\right] - \left[0\right] = 2 \log(2) - 1. \]

### Step 6: Final result
Thus, we find:

\[ \log y = 2 \log(2) - 1 \implies y = e^{2 \log(2) - 1} = \frac{4}{e}. \]

Therefore, the final answer is:

\[ \lim_{n \to \infty} \frac{1}{n} \left( (n+1)(n+2)(n+3) \cdots (n+n) \right)^{\frac{1}{n}} = \frac{4}{e}. \]