Almost Harmonic

This type of series is very close to the Kempner series. The answer is essentially already given in this wikipedia article

Here are the details. Call $S$ the serie to evaluare. Write the number in base $10^4$ instead of base 10. And look at the new series $S'$ which is the harmonic sum of all number which do not contain the "digit in base $10^4$ " which represents the number [in base 10] 2018. This series does skip numbers [in base 10] of the form ####2018 and 2018#### (where # is some base-10 digit) but it does not skip numbers of the form ##2018## or #2018### or ###2018#. Since it skips less numbers, it is larger than $S$ . So we will show that $S'$ converges and conclude that $S$ converges.

Now to estimate the series $S'$ , we work mostly in base $10^4$ . As a warm up, there are $(10^4-1) \cdot (10^4)^{k-1}$ numbers which have $k$ digits (in base $10^4$ ): indeed there are $10^4-1$ choices for the first digit (it cannot be 0) and $10^4$ choices for the remaining $k-1$ digits.

Of interest to us now is that there are $(10^4-2) \cdot (10^4-1)^{k-1}$ which do not contain the digit (in base $10^4$ !) representing the number (written in base 10) "2018": indeed there are $10^4-2$ choices for the first digit (it cannot be 0 or "2018") and $10^4-1$ choices for the remaining $k-1$ digits (they cannot be equal to "2018"). On the other hand, all these numbers are $\geq (10^4)^{k-1}$ .

This means that all the numbers with $k$ digits contribute to at most $(10^4-2) \cdot (10^4-1)^{k-1} \cdot \frac{1}{(10^4)^{k-1}}$ in the series.

Now we sum over all $k$ : $S' \leq \sum_{k \geq 1} (10^4-2) \cdot (10^4-1)^{k-1} \cdot \frac{1}{(10^4)^{k-1}} = (10^4-2) \cdot \sum_{k \geq 1} \big( \frac{10^4-1}{10^4} \big)^{k-1} = (10^4-2) \cdot \frac{1}{1- \frac{10^4-1}{10^4} }$

This shows that $S'$ is convergent. Since $S$ is smaller than $S'$ it is convergent too.