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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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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.

Aster
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    Shoring that a specific number is transcendental is usually very hard. – DanielWainfleet May 21 '22 at 06:48
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    It's not at all clear to me that this number is even irrational, but maybe I'm missing something. Do you know whether the number is irrational? If so, then you should include this (with proof or reference to a proof), and if not, then it seems to me that the question of irrationality is a better question. (If you revise your question to ask about irrationality, don't drop the transcendental part, as that would be changing your question too much I think.) Incidentally, I very much doubt estimates of its value will be of help in these questions -- I can't think of any numbers where it would. – Dave L. Renfro May 21 '22 at 07:36
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    I can't think of any numbers where it would --- I suppose with an approximation one could use an inverse symbolic calculator to obtain a guess as to what irrational/transcendental number your number might be and then try to prove equality between your number and that irrational/transcendental number. – Dave L. Renfro May 21 '22 at 08:17
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    The analysis with the algdep-command with PARI/GP reveals no sign that the number is algebraic, let alone rational. Maybe, Roth's theorem helps to prove the transcendentality , but I have not even an idea for an irrationality proof. – Peter May 21 '22 at 17:57

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