Terminating Decimals Sum

Given an integer $k$ , the decimal representation of $\frac{1}{k}$ terminates if and only if the only prime factors of $k$ are $2$ and $5$ .

Thus, we must compute $\begin{aligned} \sum_{m,n\in\mathbb{N}} \frac{1}{2^n5^m} &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{1}{2^n5^m} \\ &= \sum_{n=0}^\infty \frac{1}{2^n} \sum_{m=0}^\infty \frac{1}{5^m} && \color{#3D99F6} \text{The inner sum is a geometric series!} \\ &= \sum_{n=0}^\infty \frac{1}{2^n} \cdot \frac{5}{4} \\ &= \frac{5}{4} \sum_{n=0}^\infty \frac{1}{2^n} && \color{#3D99F6} \text{Another geometric series.} \\ &= \frac{5}{4}\cdot 2 \\ &= \frac{5}{2} = \boxed{2.5} \end{aligned}$