Not Geometric, Not Arithmetic...

When writing it out, we can see that ${ S }_{ \infty } = \frac { 1 }{ 2 } +\frac { 3 }{ 8 } +\frac { 5 }{ 32 } ...$

From this $4{ S }_{ \infty }=\frac { 4 }{ 2 } +\frac { 12 }{ 8 } +\frac { 20 }{ 32 } ...=2+(\frac { 3 }{ 2 } +\frac { 5 }{ 8 } ...)$ .

We multiply ${ S }_{ \infty }$ by 4 to "shift" the numbers so that we can subtract the two equations easily.

${ S }_{ \infty } = \frac { 1 }{ 2 } +\frac { 3 }{ 8 } +\frac { 5 }{ 32 } ...$

$4{ S }_{ \infty }=2+(\frac { 3 }{ 2 } +\frac { 5 }{ 8 } ...)$

We subtract the numbers diagonally with common denominators... To avoid negative signs I will subtract the first from the second...

$3{ S }_{ \infty }=\left( 2 \right) +\left( \frac { 3 }{ 2 } -\frac { 1 }{ 2 } \right) +\left( \frac { 5 }{ 8 } -\frac { 3 }{ 8 } \right) ...=3{ S }_{ \infty }=\frac { 2 }{ 1 } +\frac { 2 }{ 2 } +\frac { 2 }{ 8 } ...$

This is clearly a geometric infinite series plus the constant 2 at the beginning...

$3{ S }_{ \infty }=\left( 2 \right) +\left( \frac { 3 }{ 2 } -\frac { 1 }{ 2 } \right) +\left( \frac { 5 }{ 8 } -\frac { 3 }{ 8 } \right) ...=3{ S }_{ \infty }=(\frac { 2 }{ 1 } )+(\frac { 2 }{ 2 } +\frac { 2 }{ 8 } ...)=2+\frac { 1 }{ 1-\frac { 1 }{ 4 } } =2+\frac { 1 }{ \frac { 3 }{ 4 } } =\frac { 6 }{ 3 } +\frac { 4 }{ 3 } =\frac { 10 }{ 3 }$

and thus $3{ S }_{ \infty }=\frac { 10 }{ 3 }$

${ S }_{ \infty }=\frac { 10 }{ 9 }$

$10 + 9 = 19$

Q.E.D.

Another approach is as follows. The sum can be rewritten as

$S_{\infty} = \sum_{n=0}^\infty (n*(\frac{1}{4})^{n}) + (\frac{1}{2})*\sum_{n=0}^\infty (\frac{1}{4})^{n}$ .

Now note that for $|x| \lt 1$ we have that

$\sum_{n=0}^\infty x^{n} = \frac{1}{1 - x}$ .

If we differentiate both sides with respect to $x$ , (the LHS term by term), we find that

$\sum_{n=0}^\infty n*x^{n-1} = \frac{1}{(1 - x)^{2}} \Longrightarrow \sum_{n=0}^\infty n*x^{n} = \frac{x}{(1 - x)^{2}}$ .

With $x = \frac{1}{4}$ we then find that

$S_{\infty} = \dfrac{\frac{1}{4}}{(1 - \frac{1}{4})^{2}} + (\frac{1}{2})*\dfrac{1}{1 - \frac{1}{4}} = (\frac{1}{4})(\frac{16}{9}) + (\frac{1}{2})(\frac{4}{3}) = \dfrac{4}{9} + \dfrac{2}{3} = \dfrac{10}{9}$ .

Thus $A = 10, B = 9$ and $A + B = \boxed{19}$ .