If the arithmetic mean of the product of all pairs of positive integers whose sum is n is `A_n` then `lim_(n->oo) n^2/(A_n)` equals to

Answer

To solve the problem, we need to find the limit of \( \frac{n^2}{A_n} \) as \( n \) approaches infinity, where \( A_n \) is the arithmetic mean of the product of all pairs of positive integers whose sum is \( n \).

### Step-by-Step Solution:

1. **Define the pairs**:
Let the two positive integers be \( k \) and \( n-k \). The pairs will be formed by varying \( k \) from 1 to \( n-1 \).

2. **Calculate the product**:
The product of each pair is \( k(n-k) \).

3. **Sum of products**:
We need to find the sum of \( k(n-k) \) for \( k \) from 1 to \( n-1 \):
\[ S = \sum_{k=1}^{n-1} k(n-k) = \sum_{k=1}^{n-1} (nk - k^2) = n\sum_{k=1}^{n-1} k - \sum_{k=1}^{n-1} k^2 \]

4. **Use summation formulas**:
- The formula for the sum of the first \( m \) integers is:
\[ \sum_{k=1}^{m} k = \frac{m(m+1)}{2} \]
- The formula for the sum of the squares of the first \( m \) integers is:
\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} \]

5. **Apply the formulas**:
For \( m = n-1 \):
\[ \sum_{k=1}^{n-1} k = \frac{(n-1)n}{2} \]
\[ \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2(n-1)+1)}{6} = \frac{(n-1)n(2n-1)}{6} \]

6. **Substituting back into the sum**:
\[ S = n \cdot \frac{(n-1)n}{2} - \frac{(n-1)n(2n-1)}{6} \]

7. **Simplifying the expression**:
\[ S = \frac{n^2(n-1)}{2} - \frac{(n-1)n(2n-1)}{6} \]
To combine these, find a common denominator (which is 6):
\[ S = \frac{3n^2(n-1)}{6} - \frac{(n-1)n(2n-1)}{6} = \frac{(n-1)n(3n - (2n - 1))}{6} = \frac{(n-1)n(n+1)}{6} \]

8. **Calculate \( A_n \)**:
The arithmetic mean \( A_n \) is given by:
\[ A_n = \frac{S}{n-1} = \frac{(n-1)n(n+1)}{6(n-1)} = \frac{n(n+1)}{6} \]

9. **Finding the limit**:
We need to evaluate:
\[ \lim_{n \to \infty} \frac{n^2}{A_n} = \lim_{n \to \infty} \frac{n^2}{\frac{n(n+1)}{6}} = \lim_{n \to \infty} \frac{6n^2}{n(n+1)} = \lim_{n \to \infty} \frac{6n}{n+1} \]
As \( n \to \infty \), this simplifies to:
\[ \lim_{n \to \infty} \frac{6n}{n+1} = \lim_{n \to \infty} \frac{6}{1 + \frac{1}{n}} = 6 \]

### Final Answer:
\[ \lim_{n \to \infty} \frac{n^2}{A_n} = 6 \]