This is a variant of the Kempner series. To show it converges, try to find a bound on partial sums containing m digits.
That is, if unless contains the digit 4, then we can say that , so if you let , then we find that .
I think the point is that given natural numbers with digits, of them have no 4ís, and each of their reciprocals is less than .
For example, if you consider all 2-digit numbers, the 9 numbers must be removed as well as 9 more ( ). Since there are a total of 90 numbers from 10 to 99, and 18 of them contain the digit 4, only 72 of them need to be considered. Moreover, each of these has a reciprocal less than the reciprocal of the least (which is 1/10). Thus, using the notation of my previous post, we find that
It can be shown (this is your job) that in general, .