If the sum of the series

$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$

$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-\frac{1}{2 \cdot 3^{3}}+\frac{1}{3^{4}}\right)+\ldots$

is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to __________.

We can rewrite the given series as follows : <br/><br/>$$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$$ <br/><br/>The first few terms of the series are : <br/><br/>$$S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right) - \left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{18} + \frac{1}{36} + \ldots\right) - \left(\frac{1}{27} + \frac{1}{54} + \ldots\right) + \ldots$$ <br/><br/>We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$: <br/><br/>$$S = \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{2}} + \frac{\frac{1}{9}}{1-\frac{1}{2}} - \frac{\frac{1}{27}}{1-\frac{1}{2}} + \ldots$$ <br/><br/>The series can be rewritten as : <br/><br/>$$S = 2 \left(\frac{1}{{2}} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)$$ <br/><br/>Now, we can simplify and rewrite the series inside the parentheses as: <br/><br/>$$S = 2 \left[\frac{1}{{2}} + (- \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)]$$ <br/><br/>The series inside the parentheses is an infinite geometric series with the first term $a = - \frac{1}{3}$ and the common ratio $r = -\frac{1}{3}$: <br/><br/>$$S = 2 \left(\frac{1}{{2}} +\frac{a}{1-r}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{1+\frac{1}{3}}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{\frac{4}{3}}\right) $$ <br/><br/>= $2\left( {{1 \over 2} - {1 \over 3} \times {3 \over 4}} \right)$ <br/><br/>$$ = 2\left( {{1 \over 2} - {1 \over 4}} \right) = 2\left( {{1 \over 4}} \right) = {1 \over 2}$$ <br/><br/>Thus, the sum of the series is $\frac{1}{2}$, and $\alpha = 1$ and $\beta = 2$ are co-prime. <br/><br/>Therefore, $\alpha + 3\beta = 1 + 6 = 7$