I can't wrap my head around this exercise:
Show that the rational number $\frac 94$ has two different decimal expansions, namely $2.2500000\dots$ and $2.2499999\dots$ by writing these decimal expansions in the form
$$\sum_{i=0}^\infty a_i\times 10^{-i}$$
for certain numbers $$a_i$$ and then using the formula for the sum of the geometric series.
I started by defining $$a_{ix} = 0.a_{i0}a_{i1}...$$ where $$a_i = \{2,5,0\}$$ with $$a_{i4} < a_{i0} = a_{i1} < a_{i3} = a_{in}$$ for the first case, and where $$a_i = \{ 2,4,9 \}$$ with $$a_{i0} = a_{i1} < a_{i2} < a_{i3} = a_{in}$$ for the second case.
Then defining
$$\sum_{i=0}^\infty a_i\times 10^{-i}$$
as
$$\sum_{i=0}^\infty a_i\times 10^{-i} = \sum_{i=0}^n \frac{9 \times a_i}{1 - 10^{n + 1}}$$
any hints if this is on the right track?
By formula for the sum of the geometric series ${a \over 1 - r}$:
\begin{align*} 2.2499999\ldots &= 2 + \frac{2}{10^1} + \frac{4}{10^2} + \frac{9}{10^3} + \frac{9}{10^4} + \ldots \\ &= 2 + \frac{2}{10^1} + \frac{4}{10^2} + \frac{1}{10^3} \sum_{k=0}^{\infty} 9 \cdot 10^{-k} \\ &= 2 + \frac{2}{10^1} + \frac{4}{10^2} + \frac{1}{10^3} \cdot \frac{9}{1 - \frac{1}{10}} \\ &= 2.24 + \frac{1}{10^2} \cdot 1 \\ &= 2.25 \end{align*}