It is not very surprising that the sum is finite, since numbers without a 7 (or any other digit) get rarer and rarer as the number of digits increases.
Here's a proof.
Let $S$ be the harmonic series with all terms whose denominator contains the digit $k$ removed. We can write $S =S_1 + S_2 + S_3 + \ldots$, where $S_i$ is the sum of all terms whose denominator contains exactly $i$ digits, all different from $k$.
Now, the number of $i$-digit numbers that do not contain the digit $k$ is $8\cdot9^{i-1}$ (there are $8$ choices for the first digit, excluding $0$ and $k$, and $9$ choices for the other digits). [Well, if $k=0$ there are $9$ choices for the first digit, but the proof still works.] So there are $8\cdot9^{i-1}$ numbers in the sum $S_i$.
Now each number in $S_i$ is of the form $\frac1a$, where $a$ is an $i$-digit number. So $a \geq 10^{i-1}$, which implies $\frac1a \leq \frac1{10^{i-1}}$.
Therefore $S_i \leq 8\cdot\dfrac{9^{i-1} }{10^{i-1}} = 8\cdot\left(\frac9{10}\right)^{i-1}$.
So $S= \sum S_i \leq \sum 8\cdot\left(\frac9{10}\right)^{i-1}$
which is a geometric series of ratio $\frac9{10} < 1$, which converges. Since $S$ is a positive series bounded above by a converging series, $S$ converges.