Proof without words: \(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots=1
\)
Explanation
We can extract from the next figure the following infinite sum
\[\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots
\]
The same thing for the next figure
\[\frac{2}{8}+\frac{2}{32}+\frac{2}{128}+\frac{2}{512}+\cdots\]
Now, let’s focus on the next figure to find out the result we get
\[\begin{aligned}
&\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots\right)+\left(\frac{2}{8}+\frac{2}{32}+\frac{2}{128}+\frac{2}{512}+\cdots\right)=1 \\
&\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots\right)+\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\cdots\right)=1 \\
&\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\cdots=1 \\
&\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+\frac{1}{2^{6}}+\frac{1}{2^{7}}+\frac{1}{2^{8}}+\cdots=1
\end{aligned}\]
