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Prove that $.0222\ldots$ (base 3) $= .1$ (base 3) $= \frac{1}{3}$ (base 10).

First, we will show $.0222\ldots$ (base 3) $= \frac{1}{3}$ (base 10).

\begin{alignat*}{3}

2\left(\frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots\right) & = & 2\sum_{n = 2}^{\infty}\left(\frac{1}{3}\right)^n\\

& = & \frac{2}{9}\sum_{n = 0}^{\infty}\left(\frac{1}{3}\right)^n\\

& = & \frac{2}{9}\frac{1}{1 - \frac{1}{3}}\\

& = & \frac{1}{3}

\end{alignat*}

I am having trouble with the second part.

First, we will show $.0222\ldots$ (base 3) $= \frac{1}{3}$ (base 10).

\begin{alignat*}{3}

2\left(\frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots\right) & = & 2\sum_{n = 2}^{\infty}\left(\frac{1}{3}\right)^n\\

& = & \frac{2}{9}\sum_{n = 0}^{\infty}\left(\frac{1}{3}\right)^n\\

& = & \frac{2}{9}\frac{1}{1 - \frac{1}{3}}\\

& = & \frac{1}{3}

\end{alignat*}

I am having trouble with the second part.

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