The sum converges for $b=2$ and diverges for $b \geq 3$. We first consider $b \geq 3$. Suppose the sum converges; then the fact that $f(n) = n f(d)$ whenever $b^{d-1} \leq n \leq b^{d} - 1$ yields $$\sum_{n=1}^\infty \frac{1}{f(n)} = \sum_{d=1}^\infty \frac{1}{f(d)} \sum_{n=b^{d-1}}^{b^d - 1} \frac{1}{n}.$$ However, by comparing the integral of $1/x$ with a Riemann sum, we see that $$\sum_{n=b^{d-1}}^{b^d - 1} \frac{1}{n} > \int_{b^{d-1}}^{b^d} \frac{dx}{x} = \log (b^d) - \log (b^{d-1}) = \log b,$$ where $\log$ denotes the natural logarithm. Thus the sum diverges for $b \geq 3$. For $b=2$, we have a slightly different identity because $f(2) \neq 2 f(2)$. Instead, for any positive integer $i$, we have $$\sum_{n=1}^{2^i-1} \frac{1}{f(n)} = 1 + \frac{1}{2} + \frac{1}{6} + \sum_{d=3}^i \frac{1}{f(d)} \sum_{n=2^{d-1}}^{2^d - 1} \frac{1}{n}.$$ Again comparing an integral to a Riemann sum, we see that for $d\geq 3$, $$\sum_{n=2^{d-1}}^{2^d - 1} \frac{1}{n} < \frac{1}{2^{d-1}} - \frac{1}{2^d} + \int_{2^{d-1}}^{2^d} \frac{dx}{x} = \frac{1}{2^d} + \log 2 \leq \frac{1}{8} + \log 2 < 1.$$ Put $c = \frac{1}{8} + \log 2$ and $L = 1+\frac{1}{2} + \frac{1}{6(1-c)}$. Then we can prove that $\sum_{n=1}^{2^i-1} \frac{1}{f(n)} < L$ for all $i \geq 2$ by induction on $i$. The case $i=2$ is clear. For the induction, note that $$\sum_{n=1}^{2^i-1} \frac{1}{f(n)} < 1 + \frac{1}{2} + \frac{1}{6} + c \sum_{d=3}^i \frac{1}{f(d)} < 1 + \frac{1}{2} + \frac{1}{6} + c \frac{1}{6(1-c)} = 1 + \frac{1}{2} + \frac{1}{6(1-c)} = L,$$ as desired. We conclude that $\sum_{n=1}^\infty \frac{1}{f(n)}$ converges to a limit less than or equal to $L$.