Sum of Reciprocals in Base 10 with Zeroes Removed

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Theorem

The infinite series

$\ds \sum_{\map P n} \frac 1 n$

where $\map P n$ is the propositional function:

$\forall n \in \Z_{>0}: \map P n \iff$ the decimal representation of $n$ contains no instances of the digit $0$

converges to the approximate limit $23 \cdotp 10345 \ldots$


Proof

For each $k \in \N$, there are $9^k$ $k$-digit numbers containing no instances of the digit $0$.

Each of these numbers is at least $10^{k - 1}$.

Hence the reciprocals of each of these numbers is at most $\dfrac 1 {10^{k - 1}}$.

Thus:

\(\ds \sum_{\map P n} \frac 1 n\) \(<\) \(\ds \sum_{k \mathop = 1}^\infty \frac {9^k} {10^{k - 1} }\)
\(\ds \) \(=\) \(\ds \frac 9 {1 - \frac 9 {10} }\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds 90\)

showing that the sum converges.


This needs considerable tedious hard slog to complete it.
In particular: Finer approximations can be obtained (e.g. in virtue of Closed Form for Triangular Numbers/Direct Proof), but due to the slow growth of the harmonic series, many numbers must to summed to obtain the approximation in the theorem. Case in point: $H_{5 \times 10^9} < 23$
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Sources

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