Does it converge?

$S=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 2 }{ 8 } +\frac { 3 }{ 16 } +\frac { 5 }{ 32 } +\frac { 8 }{ 64 } ...\\ S=\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } +\frac { 2 }{ 4 } +\frac { 3 }{ 8 } +\frac { 5 }{ 16 } +\frac { 8 }{ 32 } ...)\\ 2S=1+\frac { 1 }{ 2 } +\frac { 2 }{ 4 } +\frac { 3 }{ 8 } +\frac { 5 }{ 16 } +\frac { 8 }{ 32 } ...\\ 2S-S=1+\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 2 }{ 16 } +\frac { 3 }{ 32 } ...\\ S=1+\frac { 1 }{ 2 } (\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 2 }{ 8 } +\frac { 3 }{ 16 } +\frac { 5 }{ 32 } +\frac { 8 }{ 64 } ..)\\ S=1+\frac { S }{ 2 } \\ \frac { S }{ 2 } =1\\ S=2$

We can also use generating functions for this problem.

Letting x = $\frac{1}{2}$

$S = x + x^{2} + 2 x^{3} + 3 x^{4} + 5 x^{5} + 8 x^{6} + 13 x^{7} + 21 x^{8} + 34 x^{9} + ...$

$x * S = 0 x + 1 x^{2} + 1 x^{3} + 2 x^{4} + 3 x^{5} + 5 x^{6} + 8 x^{7} + 13 x^{8} + 21 x^{9} + ...$

$x^{2} * S = 0 x + 0 x^{2} + 1 x^{3} + 1 x^{4} + 2 x^{5} + 3 x^{6} + 5 x^{7} + 8 x^{8} + 13 x^{9} + ...$

Now: $(1 - x - x^{2}) S = x$

$S = \frac{x}{1 - x - x^{2}}$

Substituting x = $\frac{1}{2}$ , we get S = $2$

$S = \displaystyle \sum_{n=1}^\infty (\frac{1}{\sqrt{5}} \cdot (\frac{1 + \sqrt{5}}{4})^n - \frac{1}{\sqrt{5}} \cdot (\frac{1 - \sqrt{5}}{4})^n)$

Let $\varphi_1 = \frac{1 + \sqrt{5}}{4}$ and $\varphi_2 = \frac{1 - \sqrt{5}}{4}$

Then $S = \frac{1}{\sqrt{5}} \cdot (\frac{\varphi_1}{(1 - \varphi_1)} - \frac{\varphi_2}{(1 - \varphi_2)}) = \frac{1}{\sqrt{5}} \cdot (\frac{1 + \sqrt{5}}{3 - \sqrt{5}} - \frac{1 - \sqrt{5}}{3 + \sqrt{5}})$

Therefore $S = \frac{1}{\sqrt{5}} \cdot ({2 + \sqrt{5}} - (2 - \sqrt{5})) = \boxed{2}$