An interesting series

Yes your argument seems fine.

Another argument is:

Divide the integers into blocks of 10

$$[1 \dots 10] [11 \dots 20] [21 \dots 30] \dots$$

Each block will have an number with digit $7$ in it and so the series has sum at least

$$\frac{1}{10} + \frac{1}{20} + \dots + \frac{1}{10n} + \dots $$

$$ = \frac{1}{10}(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} + \dots)$$

which is divergent.

Yet another argument which uses the following useful result (the main reason for posting my answer):

If $A = \{a_i\}$ is a sequence of natural numbers such that $\sum\limits_{i=1}^{\infty} \frac{1}{a_i}$ converges then the natural density of $A$ is zero. (This is an exercise in Ivan Niven's book on Number theory and for a proof see here: If $A\subseteq\mathbb N$ and $\sum\limits_{a\in A}\frac1a$ converges then $A$ has natural density $0$).

In our case the density of the numbers under consideration is at least $\frac{1}{10}$ and thus the sum cannot be convergent.

Your argument seems fine to me... so I'll give the argument that popped into my head when I read the question.

Sum phi(n)/n for n congruent to 7 (mod 10) and multiply by 10 (which does not affect divergence/convergence). Note that

10/7  > 1/7 +1/8 +...+1/16
10/17 > 1/17+1/18+...+1/26

and so on. The result is "larger" than the harmonic series minus the first six terms (which diverges).

Although, I actually prefer the OP's proof since it's self-contained (i.e. doesn't require the harmonic series).

Here's another proof that your sum diverges.

Consider the sum $\sum_{n-1}^\infty (1-\phi(n))/n$. This is the sum of the reciprocals of integers which don't have a 7 in their decimal expansion.

The number of integers $n$ with $1-\phi(n)=1$ and $1 \le n < 10^k$ is $9^k - 1$. (We can choose each of $k$ decimal digits to be anything but 7, except we don't want to choose all zeroes.) Thus the numbers of integers $n$ with $1-\phi(n) = 1$ and $10^{k-1} \le n < 10^k$ is $(9^k - 1) - (9^{k-1}-1) = 8 \times 9^{k-1}$.

Therefore $$ S_k = \sum_{n=10^{k-1}}^{10^k - 1} {1-\phi(n) \over n} $$ has $8 \times 9^{k-1}$ nonzero terms; they are each at most $10^{k-1}$. So $S_k \le 8 \times (9/10)^{k-1}$.

The infinite sum is then $$ \sum_{k=1}^\infty S_k \le 8 \times \sum_{k=1}^\infty (9/10)^{k-1} = 8 \times 10 = 80 $$ and in particular it's finite.

Now, the harmonic series $\sum_{n=1}^\infty 1/n$ diverges; removing terms whose sum converges (like the sum that I just showed to converge) won't change that. So your series diverges.