Originally Posted by

**aman_cc** A few related question please -

1. Is there a easy way to show if the result is rational or irrational?

Yes.

Because of the formula

$\displaystyle \prod_{n=1}^\infty (1+q^n)=1+\sum_{n=1}^\infty (-1)^n (q^{\frac{n(3n-1)}{2}}+q^{\frac{n(3n+1)}{2}})$,

we have $\displaystyle \alpha=\prod_{n=1}^\infty \left(1+\frac{1}{10^n}\right) = $ $\displaystyle 1.00001010000000000000010001\ldots - 0.11000000000100100\ldots$ (only ones and zeros).

Note as well that $\displaystyle 0.111111\cdots = \frac{1}{9}$. Therefore,

$\displaystyle \alpha + \frac{1}{9} = 1.0011212111101101111112111211\cdots$.

There may be mistakes, but you see the idea: $\displaystyle \alpha+\frac{1}{9}$ has a decimal expansion with only 0,1,2's. And because the exponents in the first formula are quadratic (not linear), this expansion is **not eventually periodic**.

We conclude that $\displaystyle \alpha+\frac{1}{9}$ is irrational, and thus so is $\displaystyle \alpha$.