Fix an integer $ b \geq 2$. Let $ f(1) \equal{} 1$, $ f(2) \equal{} 2$, and for each $ n \geq 3$, define $ f(n) \equal{} n f(d)$, where $ d$ is the number of base-$ b$ digits of $ n$. For which values of $ b$ does
\[ \sum_{n\equal{}1}^\infty \frac{1}{f(n)} 
\]
converge?
1. **Define the function \( f(n) \) and the problem setup:**
   - For an integer \( b \geq 2 \), we have the sequence \( f(n) \) defined by:
     \[
     f(1) = 1, \quad f(2) = 2, \quad \text{and for } n \geq 3, \quad f(n) = n f(d),
     \]
     where \( d \) is the number of base-\( b \) digits of \( n \).

2. **Estimate the sum \( \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{n} \):**
   - Consider the integral approximation:
     \[
     \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{n} > \int_{b^{k-1}}^{b^k} \frac{1}{x} \, dx = \ln(b^k) - \ln(b^{k-1}) = \ln b.
     \]
   - For \( b \geq 3 \), we have \( \ln b > 1 \).

3. **Upper bound for the sum \( \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{n} \):**
   - Note that:
     \[
     \frac{1}{nb} + \frac{1}{nb + 1} + \cdots + \frac{1}{nb + b - 1} < \frac{1}{n}.
     \]
   - Therefore, \( \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{n} \) is a strictly decreasing function of \( k \).

4. **Case \( b \geq 3 \):**
   - We analyze the series:
     \[
     \sum_{n = 1}^{\infty} \frac{1}{f(n)} = \sum_{k = 1}^{\infty} \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{f(n)}.
     \]
   - Using the previous estimates:
     \[
     \sum_{n = b^{k-1}}^{b^k - 1} \frac{1}{f(n)} > \ln b \sum_{k = 1}^{\infty} \frac{1}{f(k)}.
     \]
   - Since \( \ln b > 1 \) for \( b \geq 3 \), the series diverges.

5. **Case \( b = 2 \):**
   - We analyze the series:
     \[
     \sum_{n = 1}^{\infty} \frac{1}{f(n)} = 1 + \frac{1}{2} + \frac{1}{6} + \sum_{n = 4}^{\infty} \frac{1}{f(n)}.
     \]
   - For \( k \geq 3 \):
     \[
     \sum_{n = 2^{k-1}}^{2^k - 1} \frac{1}{n} \leq \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} = \frac{319}{420} < 1.
     \]
   - Therefore:
     \[
     \sum_{n = 3}^{\infty} \frac{1}{f(n)} < \frac{1}{6} + \left( \frac{319}{420} \right) \sum_{k = 3}^{\infty} \frac{1}{f(k)}.
     \]
   - This implies:
     \[
     \left( \frac{101}{420} \right) \sum_{n = 3}^{\infty} \frac{1}{f(n)} < \frac{1}{6},
     \]
     and thus the series converges.

The final answer is \( \boxed{ b = 2 } \).