To solve the problem, we start by analyzing the given recurrence relation and using reciprocal transformation.

Step 1: Reciprocal Transformation

Given (a1 = 1) and (a{n+1} = \frac{a_n}{1 + n an}), take reciprocals of both sides:
[ \frac{1}{a{n+1}} = \frac{1 + n a_n}{a_n} = \frac{1}{a_n} + n ]

Let (b_n = \frac{1}{an}), then the recurrence becomes:
[ b{n+1} = b_n + n ]

Step 2: Compute (b_n)

Since (b_1 = \frac{1}{a_1} = 1), we expand the recurrence:
[ b_n = b1 + \sum{k=1}^{n-1} k ]

The sum (\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}), so:
[ b_n = 1 + \frac{n(n-1)}{2} = \frac{2 + n(n-1)}{2} = \frac{n^2 - n + 2}{2} ]

Step 3: Find (a_n)

[ a_n = \frac{1}{b_n} = \frac{2}{n^2 - n + 2} ]

Answer: (\boxed{\dfrac{2}{n^2 - n + 2}})

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