2022-02-27

How do you prove that
$\frac{{10}^{n}-1}{{10}^{n}-2}=1+\sum _{i=0}^{\mathrm{\infty }}{2}^{i}×{10}^{-n\left(i+1\right)}$

Elijah Hunt

Your sum is a geometric series:
$\sum _{i=0}^{\mathrm{\infty }}{2}^{i}×{10}^{-n\left(i+1\right)}={10}^{-n}\sum _{i=0}^{\mathrm{\infty }}{\left(\frac{2}{{10}^{n}}\right)}^{i}$
$={10}^{-n}×\frac{1}{1-2×{10}^{-n}}=\frac{1}{{10}^{n}-2}$
You are done once you observe that
$\frac{{10}^{n}-1}{{10}^{n}-2}=\frac{{10}^{n}-2}{{10}^{n}-2}+\frac{1}{{10}^{n}-2}$
$=1+\frac{1}{{10}^{n}-2}$

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