For each positive integer `n` , let `y_n=1/n((n+1)(n+2)dot(n+n))^(1/n)` For `x in R` let `[x]` be the greatest integer less than or equal to `x` . If `(lim)_(n->oo)y_n=L` , then the value of `[L]` is ______.

Explanation

To solve the problem, we need to find the limit \( L \) of the sequence defined by

\[ y_n = \frac{1}{n} \left( (n+1)(n+2) \cdots (n+n) \right)^{\frac{1}{n}} \]

and then determine the greatest integer less than or equal to \( L \).

### Step 1: Rewrite \( y_n \)

We start with the expression for \( y_n \):

\[ y_n = \frac{1}{n} \left( (n+1)(n+2) \cdots (n+n) \right)^{\frac{1}{n}} \]

The product \( (n+1)(n+2) \cdots (n+n) \) can be rewritten as:

\[ (n+1)(n+2) \cdots (n+n) = \prod_{k=1}^{n} (n+k) \]

### Step 2: Take the logarithm

Taking the natural logarithm of \( y_n \):

\[ \log y_n = \log \left( \frac{1}{n} \right) + \frac{1}{n} \log \left( \prod_{k=1}^{n} (n+k) \right) \]

Using the property of logarithms, we can express this as:

\[ \log y_n = -\log n + \frac{1}{n} \sum_{k=1}^{n} \log(n+k) \]

### Step 3: Simplify the sum

We can rewrite the sum:

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

Thus, we have:

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

### Step 4: Substitute back into the logarithm

Now substituting back into the expression for \( \log y_n \):

\[ \log y_n = -\log n + \frac{1}{n} \left( n \log n + \sum_{k=1}^{n} \log\left(1 + \frac{k}{n}\right) \right) \]

This simplifies to:

\[ \log y_n = \sum_{k=1}^{n} \frac{1}{n} \log\left(1 + \frac{k}{n}\right) \]

### Step 5: Take the limit as \( n \to \infty \)

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

\[ \lim_{n \to \infty} \log y_n = \int_{0}^{1} \log(1+x) \, dx \]

### Step 6: Evaluate the integral

To evaluate the integral:

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

Using 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 integral \( \int \frac{x}{1+x} \, dx \) can be simplified and evaluated.

### Step 7: Find the limit

After evaluating the integral, we find:

\[ \lim_{n \to \infty} \log y_n = \log 4 - 1 \]

Thus,

\[ L = e^{\log 4 - 1} = \frac{4}{e} \]

### Step 8: Find the greatest integer less than or equal to \( L \)

To find \( [L] \):

\[ L = \frac{4}{e} \approx \frac{4}{2.718} \approx 1.47 \]

Thus, the greatest integer less than or equal to \( L \) is:

\[ \lfloor L \rfloor = 1 \]

### Final Answer

The value of \( [L] \) is \( \boxed{1} \).