Let $f(n)=\displaystyle\sum_{k=1}^n \dfrac{1}{k}$.  Then there exists constants $\gamma$, $c$, and $d$ such that \[f(n)=\ln(x)+\gamma+\dfrac{c}{n}+\dfrac{d}{n^2}+O\left(\dfrac{1}{n^3}\right),\] where the $O\left(\dfrac{1}{n^3}\right)$ means terms of order $\dfrac{1}{n^3}$ or lower.  Compute the ordered pair $(c,d)$.
1. **Define the Harmonic Series**:
   The function \( f(n) \) is defined as the \( n \)-th Harmonic number:
   \[
   f(n) = \sum_{k=1}^n \frac{1}{k}
   \]

2. **Approximate the Harmonic Series**:
   The Harmonic series can be approximated using the natural logarithm and the Euler-Mascheroni constant \( \gamma \):
   \[
   f(n) \approx \ln(n) + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + O\left(\frac{1}{n^3}\right)
   \]
   This approximation comes from the Euler-Maclaurin formula.

3. **Compare with Given Expression**:
   We are given the expression:
   \[
   f(n) = \ln(n) + \gamma + \frac{c}{n} + \frac{d}{n^2} + O\left(\frac{1}{n^3}\right)
   \]
   By comparing the two expressions, we can identify the constants \( c \) and \( d \).

4. **Identify Constants**:
   From the approximation:
   \[
   f(n) \approx \ln(n) + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + O\left(\frac{1}{n^3}\right)
   \]
   We see that:
   \[
   c = \frac{1}{2}
   \]
   and
   \[
   d = -\frac{1}{12}
   \]

5. **Conclusion**:
   The ordered pair \( (c, d) \) is:
   \[
   (c, d) = \left( \frac{1}{2}, -\frac{1}{12} \right)
   \]

The final answer is \( \boxed{ \left( \frac{1}{2}, -\frac{1}{12} \right) } \)