Find the positive constant $c_0$ such that the series \[ \displaystyle\sum_{n = 0}^{\infty} \dfrac {n!}{(cn)^n} \] converges for $c>c_0$ and diverges for $0<c<c_0$.
To determine the positive constant \( c_0 \) such that the series 
\[ \sum_{n = 0}^{\infty} \frac{n!}{(cn)^n} \]
converges for \( c > c_0 \) and diverges for \( 0 < c < c_0 \), we can use Stirling's approximation and the ratio test.

### Using Stirling's Approximation

1. **Stirling's Approximation**:
   Stirling's approximation states that 
   \[ n! \approx \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{e^n} \]
   as \( n \to \infty \).

2. **Rewrite the Series Term**:
   Using Stirling's approximation, the term of the series becomes:
   \[ \frac{n!}{(cn)^n} \approx \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{(cn)^n} = \frac{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{c^n n^n} = \frac{\sqrt{2\pi n}}{(ec)^n} \]

3. **Evaluate the Limit**:
   To determine convergence, we need to evaluate the limit:
   \[ \lim_{n \to \infty} \frac{\sqrt{2\pi n}}{(ec)^n} \]
   For the series to converge, this limit must be zero:
   \[ \lim_{n \to \infty} \frac{\sqrt{2\pi n}}{(ec)^n} = 0 \]
   This happens if and only if \( ec > 1 \), or equivalently:
   \[ c > \frac{1}{e} \]

### Using the Ratio Test

1. **Ratio Test**:
   The ratio test states that a series \( \sum a_n \) converges if:
   \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \]

2. **Apply the Ratio Test**:
   Let \( a_n = \frac{n!}{(cn)^n} \). Then:
   \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(c(n+1))^{n+1}} \cdot \frac{(cn)^n}{n!} = \frac{(n+1) \cdot n!}{(c(n+1))^{n+1}} \cdot \frac{(cn)^n}{n!} = \frac{(n+1) \cdot (cn)^n}{(c(n+1))^{n+1}} \]

3. **Simplify the Expression**:
   \[ \frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot (cn)^n}{c^{n+1} (n+1)^{n+1}} = \frac{(cn)^n}{c^{n+1} (n+1)^n} = \frac{1}{c} \cdot \left( \frac{n}{n+1} \right)^n \]

4. **Evaluate the Limit**:
   \[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{c} \cdot \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n \]
   Using the limit definition of \( e \):
   \[ \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n = \frac{1}{e} \]
   Therefore:
   \[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{c} \cdot \frac{1}{e} = \frac{1}{ce} \]

5. **Determine Convergence**:
   For the series to converge, we need:
   \[ \frac{1}{ce} < 1 \]
   This implies:
   \[ c > \frac{1}{e} \]

Thus, the positive constant \( c_0 \) is \( \frac{1}{e} \).

The final answer is \( \boxed{\frac{1}{e}} \)