Let `lim _( x to oo) n ^((1)/(2 )(1+(1 )/(n))). (1 ^(1) . 2 ^(2) . 3 ^(3)....n ^(n ))^((1)/(n ^(2)))=e^((-p)/(q)) `
where p and q are relative prime positive integers. Find the value of `|p+q|.`

Answer

To solve the limit problem given by

\[ \lim_{n \to \infty} n^{\frac{1}{2}\left(1 + \frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n\right)^{\frac{1}{n^2}} = e^{-\frac{p}{q}}, \]

where \( p \) and \( q \) are relatively prime positive integers, we will follow these steps:

### Step 1: Rewrite the limit expression
Let

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

### Step 2: Take the logarithm of \( L \)
Taking the natural logarithm of both sides, we have:

\[ \log L = \lim_{n \to \infty} \left( \frac{1}{2}\left(1 + \frac{1}{n}\right) \log n + \frac{1}{n^2} \log(1^1 \cdot 2^2 \cdots n^n) \right). \]

### Step 3: Simplify the logarithm of the product
The logarithm of the product can be expressed as:

\[ \log(1^1 \cdot 2^2 \cdots n^n) = \sum_{k=1}^{n} k \log k. \]

Thus,

\[ \log L = \lim_{n \to \infty} \left( \frac{1}{2}\left(1 + \frac{1}{n}\right) \log n + \frac{1}{n^2} \sum_{k=1}^{n} k \log k \right). \]

### Step 4: Analyze the first term
As \( n \to \infty \), the first term becomes:

\[ \frac{1}{2}\left(1 + \frac{1}{n}\right) \log n \to \frac{1}{2} \log n. \]

### Step 5: Analyze the second term
For the second term, we can approximate the sum using integrals:

\[ \sum_{k=1}^{n} k \log k \approx \int_1^n x \log x \, dx. \]

Calculating this integral:

\[ \int x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C. \]

Evaluating from 1 to \( n \):

\[ \left[ \frac{n^2}{2} \log n - \frac{n^2}{4} \right] - \left[ \frac{1}{2} \log 1 - \frac{1}{4} \right] = \frac{n^2}{2} \log n - \frac{n^2}{4} + \frac{1}{4}. \]

Thus,

\[ \sum_{k=1}^{n} k \log k \approx \frac{n^2}{2} \log n - \frac{n^2}{4}. \]

### Step 6: Substitute back into the limit
Now substituting back into our expression for \( \log L \):

\[ \log L = \lim_{n \to \infty} \left( \frac{1}{2} \log n + \frac{1}{n^2} \left( \frac{n^2}{2} \log n - \frac{n^2}{4} \right) \right). \]

This simplifies to:

\[ \log L = \lim_{n \to \infty} \left( \frac{1}{2} \log n + \frac{1}{2} \log n - \frac{1}{4} \right) = \lim_{n \to \infty} \log n - \frac{1}{4}. \]

### Step 7: Evaluate the limit
As \( n \to \infty \), \( \log n \to \infty \), thus:

\[ \log L \to -\frac{1}{4}. \]

### Step 8: Exponentiate to find \( L \)
Thus, we have:

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

### Step 9: Identify \( p \) and \( q \)
From the expression \( e^{-\frac{p}{q}} = e^{-\frac{1}{4}} \), we identify \( p = 1 \) and \( q = 4 \). Since \( p \) and \( q \) are relatively prime, we find:

\[ |p + q| = |1 + 4| = 5. \]

### Final Answer
The value of \( |p + q| \) is

\[ \boxed{5}. \]