AGP really?

By Trigonometric Periodicity Identities , we observe that for $x =\dfrac \pi3$ ,
$\sin(x)=\sin(7x)=\sin(13x)=\cdots=\frac{\sqrt{3}}{2}$ ,
$\sin(3x)=\sin(9x)=\cdots=0$ , and
$\sin(5x)=\sin(11x)=\cdots=\frac{-\sqrt{3}}{2}$ .

So the series becomes

$\frac{\sqrt{3}}{2}\left(1+\frac{1}{3^3}+\frac{1}{3^6}+.....-\frac{1}{3^2}-\frac{1}{3^5}...\right)=\frac{\sqrt{3}}{2}\left(\frac{27}{26}-\frac{3}{26}\right)=\frac{6\sqrt{3}}{13}$

The penultimate step can be solved by applying the Geometric Progression Sum with common ratio, $r = 1/3^3$ .

This is a rather weird approach, but I'll give it a try, anyway....

$f(x)$ is the imaginary component of the complex function $g(x) = \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{i(2n + 1)x}}{3^{n}}.$

The terms of this series are in geometric progression, as the common ratio is $\dfrac{e^{i2x}}{3},$ the absolute value of which is $\dfrac{1}{3}.$ So treating this series as a "normal" infinite geometric series, its sum will be

$g(x) = \dfrac{e^{ix}}{1 - \dfrac{e^{i2x}}{3}} = \dfrac{3e^{ix}}{3 - e^{i2x}} = \dfrac{3}{3e^{-ix} - e^{ix}}.$

Plugging in $x = \dfrac{\pi}{3},$ we end up with

$g(\frac{\pi}{3}) = \dfrac{3}{3(\cos(\frac{\pi}{3}) - i*\sin(\frac{\pi}{3})) - (\cos(\frac{\pi}{3}) + i*\sin(\frac{\pi}{3}))} =$

$\dfrac{3}{\frac{3}{2} - i*\frac{3\sqrt{3}}{2} - \frac{1}{2} - i*\frac{\sqrt{3}}{2}} = \dfrac{3}{1 - i*2\sqrt{3}} * \dfrac{1 + i*2\sqrt{3}}{1 + i*2\sqrt{3}} = \dfrac{3 + i*6\sqrt{3}}{13}.$

Since $f(x)$ is the imaginary component of $g(x),$ we see that $f(\frac{\pi}{3}) = \dfrac{6\sqrt{3}}{13},$ and so

$A + B + C = 6 + 3 + 13 = \boxed{22}.$

Here is my calculus free approach

Call the summation to be S

$\dfrac{S}{1} = \dfrac {\sin x}{1}+\dfrac{\sin 3x}{3} \cdots$

We Use The Traditional Method Of Shifting Terms in AGP

$\dfrac{S}{3} = \dfrac{\sin x}{3}+\dfrac{\sin 3x}{9} \cdots$

Subtracting These We Get

$\dfrac{2S}{3} = \sin x+\dfrac{sin 3x - sin x}{3}+\dfrac{sin 5x-sin 3x}{9} \cdots$

Applying The Relevant Trigonometric Identity

$\dfrac{2S}{3} = \sin x+ \dfrac{2\times sin x \times cos 2x}{3} +\dfrac{2\times sinx\times cos 4x}{9} \cdots$

Now From Every Term We Take 2sinx Common

$\dfrac{2S}{3}= \sin x + 2\times \sin x (\dfrac{cos 2x}{3} + \dfrac{cos 4x}{9} + \cdots$ )

Now i will evaluate the cosine series separately

Let

$\dfrac{Z}{1} = \dfrac{cos 2x}{3}+\dfrac{cos 4x}{9}+\dfrac{cos 6x}{27}+\cdots$

Again we apply the method of shifting the terms

$\dfrac{Z}{3} = \dfrac{cos 2x}{9} +\dfrac{cos 4x}{27}+\cdots$

Subtracting

$\dfrac{2Z}{3} = \dfrac{cos 2x}{3} + \dfrac{cos 4x - cos 2x}{9} + \dfrac{cos 6x -cos 4x}{27}+\cdots$

Again applying the relevant trigonometric identity

$\dfrac{2Z}{3} = \dfrac{cos 2x}{3} - \dfrac{2\times sin 3x\times sin x}{9} - \dfrac{2\times sin 5x\times sin x}{27}+\cdots$

Multiplying by 3 And taking 2sinx common

$\dfrac{2Z}{1} = \cos 2x - 2\times sin x(\dfrac{sin 3x}{3}+\dfrac{sin5x}{9}+\cdots$

And to our suprise the series inside the bracket is the series asked to us (S)

So we can write

$\dfrac{2Z}{1} = \cos 2x - 2\times sin x\times(S-\sin x)$

Also we have the relation

$\dfrac{2S}{3} = \sin x + 2\times sin x\times Z$

Solving the system of equations above simultaneously we get the value of S

Finally on simplification and applying identities

We get

$\dfrac{S}{1} = \dfrac{3\times sin x}{1+3\times\ sin^2 x}$