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- Thread starter dwsmith
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- Feb 29, 2012

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What if

$$2 \frac{1}{3} + \frac{1}{3^2} + 2 \frac{1}{3^3} + \frac{1}{3^4} + 2 \frac{1}{3^5} + \cdots = \left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \cdots \right) + \left( \frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \cdots \right)$$

$$= \frac{1}{3} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots \right) + \frac{1}{3}\left( 1 + \frac{1}{3^2} + \frac{1}{3^4} + \cdots \right).$$

The first parenthesis is a geometric series with ratio $\frac{1}{3}$ and the second with ratio $\frac{1}{9}$.

This doesn't seem right though, because the sum would be $\frac{7}{8}$, whereas on the right side you'd get $\frac{1}{2}$. Perhaps this change of terms in the series I did is wrong, but I doubt that because the series on the right, which it is supposed to be, is absolutely convergent, being a geometric series with ratio $\frac{1}{3}$.

I hope this helps spring the discussion into the right direction.

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