Throw $ n$ balls in to $ 2n$ boxes.　Suppose each ball comes into each box with equal probability of entering in any boxes.
Let $ p_n$ be the probability such that any box has ball less than or equal to one. Find the limit $ \lim_{n\to\infty} \frac{\ln p_n}{n}$
1. We start by defining the problem: We have \( n \) balls and \( 2n \) boxes. Each ball is thrown into a box with equal probability, meaning each box has a probability of \( \frac{1}{2n} \) of receiving any particular ball.

2. Let \( p_n \) be the probability that no box contains more than one ball. We need to find the limit \( \lim_{n \to \infty} \frac{\ln p_n}{n} \).

3. Define \( a_k \) as the probability that a specific box is empty after \( k \) balls have been thrown, and \( b_k \) as the probability that a specific box contains exactly one ball after \( k \) balls have been thrown.

4. For \( q = 1 - \frac{1}{2n} \), we have:
   \[
   a_k = q^k
   \]
   This is because the probability that a specific box is empty after \( k \) balls is the probability that each of the \( k \) balls did not go into that box, which is \( q^k \).

5. The probability \( b_k \) can be derived as follows:
   \[
   b_k = (1 - q) a_{k-1} + q b_{k-1}
   \]
   This recurrence relation accounts for the probability that the box has exactly one ball after \( k \) balls: either the box was empty after \( k-1 \) balls and the \( k \)-th ball went into it, or the box already had one ball and the \( k \)-th ball did not go into it.

6. Using the initial conditions \( a_1 = q \) and \( b_1 = 1 - q \), we can solve the recurrence relation:
   \[
   b_k = (1 - q) q^{k-1} + q b_{k-1}
   \]
   By iterating this, we get:
   \[
   b_k = k q^{k-1} (1 - q)
   \]

7. The probability \( p_n \) that no box contains more than one ball is given by:
   \[
   p_n = a_n + b_n = q^n + n q^{n-1} (1 - q)
   \]

8. We need to find the limit:
   \[
   \lim_{n \to \infty} \frac{\ln p_n}{n}
   \]

9. First, simplify \( p_n \):
   \[
   p_n = q^n + n q^{n-1} (1 - q)
   \]
   For large \( n \), \( q = 1 - \frac{1}{2n} \approx e^{-\frac{1}{2n}} \), so:
   \[
   q^n \approx e^{-\frac{n}{2n}} = e^{-\frac{1}{2}}
   \]
   and:
   \[
   n q^{n-1} (1 - q) \approx n e^{-\frac{n-1}{2n}} \left( \frac{1}{2n} \right) \approx n e^{-\frac{1}{2}} \frac{1}{2n} = \frac{1}{2} e^{-\frac{1}{2}}
   \]

10. Therefore, for large \( n \):
    \[
    p_n \approx e^{-\frac{1}{2}} + \frac{1}{2} e^{-\frac{1}{2}} = \frac{3}{2} e^{-\frac{1}{2}}
    \]

11. Now, compute the limit:
    \[
    \lim_{n \to \infty} \frac{\ln p_n}{n} = \lim_{n \to \infty} \frac{\ln \left( \frac{3}{2} e^{-\frac{1}{2}} \right)}{n} = \lim_{n \to \infty} \frac{\ln \frac{3}{2} + \ln e^{-\frac{1}{2}}}{n} = \lim_{n \to \infty} \frac{\ln \frac{3}{2} - \frac{1}{2}}{n} = 0
    \]

The final answer is \( \boxed{ 0 } \)