Peculiar Sums

$\phi = (1 + \sqrt{5})/2 \; \; \; , \; \; \; 1/\phi = (-1 + \sqrt{5})/2$

$A = \frac{1}{\phi} + ( \frac{1}{\phi} )^{2} + ( \frac{1}{\phi} )^{3} + ....$

This is a geometric series having $\frac{1}{\phi}$ for the first term, and each successive term is multiplied by $\frac{1}{\phi}$ . Since -1 < $\frac{1}{\phi}$ < 1 the series converges and its sum is

$A = \frac{1/\phi}{1 - \frac{1}{\phi}} \cdot \frac{\phi}{\phi} = \frac{1}{\phi - 1} = \phi$

.

$B = 1 + \frac{1}{\phi^{2}} + \frac{1}{\phi^{4}} + \frac{1}{\phi^{6}} + .....$

This is a geometric series having 1 for the first term, and each successive term is multiplied by $\frac{1}{\phi^{2}}$ . Since -1 < $\frac{1}{\phi^{2}}$ < 1 the series converges and its sum is

$B = 1/(1 - \frac{1}{\phi^{2}}) = [1 - \frac{1}{4}(6 - 2\sqrt{5})]^{-1} = [-\frac{1}{2} + \frac{1}{2} \sqrt{5}]^{-1} = [1/\phi]^{-1} = \phi$

$A + B = 2\phi$

$\phi$ is the golden mean.