Is the sum of infinite recurring decimals also a recurring decimal?

I am curious to know if $N=0.12233344444455555...$ is a rational or an irrational number. I see that, since it can be obtained by the sum of $0+0.1+0.022+0.000333+...$, it could be obtained by this infinite series:

$$N=\sum_{n=1}^\infty n\left(\sum_{m=1}^n 10^{m-1}\right)10^{-{n(n+1)}/{2}}$$

In addition, since it can also be obtained by the sum of $0.\overline{1}+0.0\overline{1}+0.000\overline{1}+...=\frac{1}{9}+\frac{1}{90}+\frac{1}{9000}+...$, it can also be obtained by a sum of infinite recurring decimals:

$$N=\sum_{n=0}^\infty \frac{1}{9·10^{{n(n+1)}/{2}}} =\frac{1}{9} \sum_{n=0}^\infty 10^{-{n(n+1)}/{2}}$$

Would this number, obtained as an infinite sum of recurring decimals, be also a recurring decimal? If not, since it is also the sum of infinite rational numbers, would it even be rational?

Solution 1:

The second way you've written the summation makes it clear the number $N$ is irrational, because $9N$ is non-terminating and non-recurring:

$$ 9N = 1.10100100001000001\ldots $$

Note that location of the digit $1$ from the $10^{-(n-1)n/2}$ term is separated from the next corresponding digit $1$ from the $10^{-n(n+1)/2}$ term by $n$ decimal places, i.e. by $n-1$ intervening zero digits. Thus the decimal expansion never becomes repeating of any finite length digit pattern. But any rational number would so repeat, and $N$ is rational if and only if $9N$ is rational.