Determine the general term of the sequence ($a_n$) given by $a_0 =\alpha > 0$ and $a_{n+1} =\frac{a_n}{1+a_n}$
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1. Given the sequence \(a_n\) defined by \(a_0 = \alpha > 0\) and \(a_{n+1} = \frac{a_n}{1 + a_n}\), we need to determine the general term \(a_n\).

2. To simplify the recurrence relation, we introduce a new sequence \(b_n\) such that \(b_n = \frac{1}{a_n}\). This transformation will help us linearize the recurrence relation.

3. Substituting \(b_n = \frac{1}{a_n}\) into the recurrence relation \(a_{n+1} = \frac{a_n}{1 + a_n}\), we get:
   \[
   a_{n+1} = \frac{a_n}{1 + a_n} \implies \frac{1}{b_{n+1}} = \frac{\frac{1}{b_n}}{1 + \frac{1}{b_n}} = \frac{1}{b_n + 1}
   \]
   Taking the reciprocal of both sides, we obtain:
   \[
   b_{n+1} = b_n + 1
   \]

4. The recurrence relation \(b_{n+1} = b_n + 1\) is a simple arithmetic progression. Solving this, we get:
   \[
   b_{n+1} - b_n = 1 \implies b_{n+1} - b_0 = n + 1 \implies b_n = b_0 + n
   \]
   Given that \(b_0 = \frac{1}{a_0} = \frac{1}{\alpha}\), we have:
   \[
   b_n = \frac{1}{\alpha} + n
   \]

5. Now, we revert back to the original sequence \(a_n\) using \(a_n = \frac{1}{b_n}\):
   \[
   a_n = \frac{1}{b_n} = \frac{1}{\frac{1}{\alpha} + n} = \frac{\alpha}{1 + \alpha n}
   \]

6. Therefore, the general term of the sequence \(a_n\) is:
   \[
   \boxed{a_n = \frac{\alpha}{1 + \alpha n}}
   \]