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Obviously:

$$\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\cdots=0.1111\dots=\frac{1}{9}$$

is a rational number.

Now, if we make terms with demoninators in the form:

$$q_n=\sum_{k=0}^{n} 10^k$$

Then the sum will be:

$$\sum_{n=1}^{\infty}\frac{1}{q_n}=\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots=0.1009181908362007\dots$$

The decimal expansion of this number appears to be non-periodic.

How can we prove/disprove that this number is irrational?

Edit

This number is at OEIS: http://oeis.org/A065444

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If we define the "Lambert series" $$ f(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n} = \sum_{n=1}^\infty \frac{1}{(1/x)^n-1}, $$ then since $q_n = (10^{n+1}-1)/9$, your number is $$ \sum_{n=1}^\infty \frac1{q_n} = 9f(\tfrac1{10})-1. $$ Chowla proved in 1947 that $f(\frac1{10})$ is irrational, and hence so is your number. You can find a proof in this 1948 paper of Erdös.

(Interesting side note: we also have $$ f(x) = \sum_{m=1}^\infty \tau(m)x^m, $$ where $\tau(m)$ is the number of positive divisors of $m$.)

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    $\begingroup$ Is it only irrational or even transcendental? $\endgroup$ – MrYouMath Apr 25 '16 at 8:43
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    $\begingroup$ @MrYouMath that's probably unknown. Proving numbers are transcendental is really hard $\endgroup$ – Stella Biderman Apr 25 '16 at 10:43
  • $\begingroup$ One could search MathSciNet or Google Scholar for "Lambert series transcendental" to see if anyone's made progress on that question. $\endgroup$ – Greg Martin Apr 25 '16 at 18:10
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This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series.

$$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$ $q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$

$\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ terms).

$$\sum_{n=0}^m \frac{1}{q_n}=\sum_{n=1}^m \frac{9}{10^{n+1}-1}= \frac{9}{\ln(10)}\left(\psi_{10}(m+2)-\psi_{10}(1)\right)-9(m+1)$$ $\psi_q(x)$ is the q-digamma function. The leading terms of the asymptotic series are: $$\psi_{10}(x)\sim (x-\frac{1}{2})\ln(10)-\ln(9)$$

$$\sum_{n=0}^{m\to\infty} \frac{1}{q_n}\sim \frac{9}{\ln(10)}\left( (m+2-\frac{1}{2})\ln(10)-\ln(9)-\psi_{10}(1)\right)-9(m+1)$$

After simplification : $$\sum_{n=0}^{\infty} \frac{1}{q_n}=\frac{9}{2} -\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)$$ $\psi_{10}(1)\simeq-1.32759401026424207 $

$\frac{9}{2} -\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)\simeq 1.100918190836200736 $

Note that $\sum_{n=0}^{\infty} \frac{1}{q_n}$ includes the first term$=1$

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