Let $k{}$ be a positive integer. For every positive integer $n \leq 3^k$, denote $b_n$ the greatest power of $3$ that divides $C_{3^k}^n$. Compute $\sum_{n=1}^{3^k-1} \frac{1}{b_n}$.
1. **Understanding the problem**: We need to compute the sum $\sum_{n=1}^{3^k-1} \frac{1}{b_n}$, where $b_n$ is the greatest power of $3$ that divides $C_{3^k}^n$ (the binomial coefficient $\binom{3^k}{n}$).

2. **Using Kummer's Theorem**: Kummer's theorem states that the exponent of a prime $p$ in the binomial coefficient $\binom{m}{n}$ is equal to the number of carries when $n$ is subtracted from $m$ in base $p$. For our problem, $p = 3$ and $m = 3^k$.

3. **Properties of binomial coefficients**: For $n$ in the range $1 \leq n \leq 3^k - 1$, the binomial coefficient $\binom{3^k}{n}$ is divisible by $3$. This is because $3^k$ is a power of $3$, and any binomial coefficient $\binom{3^k}{n}$ for $0 < n < 3^k$ will have at least one factor of $3$.

4. **Greatest power of $3$ dividing $\binom{3^k}{n}$**: The greatest power of $3$ that divides $\binom{3^k}{n}$ can be determined by counting the number of carries in the subtraction $3^k - n$ in base $3$. For each $n$, the number of carries is the number of digits in $n$ that are non-zero in base $3$.

5. **Summing the reciprocals**: We need to sum the reciprocals of these powers of $3$. For each $n$, the power of $3$ that divides $\binom{3^k}{n}$ is $3^c$, where $c$ is the number of carries. Therefore, $b_n = 3^c$ and $\frac{1}{b_n} = 3^{-c}$.

6. **Counting the carries**: For each $n$ from $1$ to $3^k - 1$, we need to count the number of carries in the subtraction $3^k - n$ in base $3$. This is equivalent to counting the number of non-zero digits in the base $3$ representation of $n$.

7. **Summing over all $n$**: We need to sum $3^{-c}$ over all $n$ from $1$ to $3^k - 1$. Each $n$ can be represented in base $3$ with $k$ digits, and the number of non-zero digits in $n$ determines the value of $c$.

8. **Final computation**: The sum $\sum_{n=1}^{3^k-1} \frac{1}{b_n}$ can be computed by considering the contribution of each possible number of non-zero digits in the base $3$ representation of $n$. For each $c$ from $1$ to $k$, there are $\binom{k}{c} \cdot 2^c$ numbers with exactly $c$ non-zero digits, and each contributes $3^{-c}$ to the sum.

\[
\sum_{n=1}^{3^k-1} \frac{1}{b_n} = \sum_{c=1}^{k} \binom{k}{c} \cdot 2^c \cdot 3^{-c}
\]

9. **Simplifying the sum**: The sum can be simplified using the binomial theorem:

\[
\sum_{c=1}^{k} \binom{k}{c} \cdot \left(\frac{2}{3}\right)^c = \left(1 + \frac{2}{3}\right)^k - 1 = \left(\frac{5}{3}\right)^k - 1
\]

Conclusion:
\[
\boxed{\left(\frac{5}{3}\right)^k - 1}
\]