Is this sum a transcendental number?

Question

Question

Is the sum $S$ a transcendental number?

$$S = 1 + \frac{1}{12}+\frac{1}{123}+\frac{1}{1234}+\cdots$$

123456789 is followed by 12345678910, 1234567891011

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Some Search Results

The denominator has a recursive relationship:

$$a_{n}=a_{n-1} \times 10^{\left\lfloor\lg(10 n)\right\rfloor }+n$$

Champernowne constant is a transcendental number.

The reciprocal of a transcendental number is also a transcendental number.

Clear[a];
RSolve[{a[n] == a[n - 1] * 10^Floor[Log10[10 * n]] + n, a[1] == 1}, a[n], n]
a[1] = 1;
a[n_] := a[n] = a[n - 1] * 10^Floor[Log10[10 * n]] + n
Table[Sum[1 / a[i], {i, 1, j}], {j, 1, 100}] // N

Numerical calculation yields $S \approx 1.092363791$, convergence is fast because each contribution decays exponentially.