Infinite Infinities

I have written the solution on a paper. Please see the image .

Since this goes on for infinity, this can be written as:

$2\sum _{ n=0 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } =x$

Now let:

$2\sum _{ n=0 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } =S$

$\sum _{ n=0 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } =\frac { S }{ 2 }$

$\sum _{ n=1 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } =\frac { S }{ 2 } \frac { 1 }{ \left[ \frac { x }{ 2013 } \right] } \\ \therefore \sum _{ n=0 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } -\sum _{ n=1 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } ={ \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ 0 } \right] }^{ -1 }=\frac { 1 }{ 2 } \\ =\frac { S }{ 2 } -\frac { S }{ 2 } \frac { 1 }{ \left[ \frac { x }{ 2013 } \right] } =\frac { Sx-2013S }{ 2x }$

Solving for $S$ gives: $S=\frac { x }{ x-2013 }$

And since $2\sum _{ n=0 }^{ \infty }{ { \left[ { 2\left[ \frac { x }{ 2013 } \right] }^{ n } \right] }^{ -1 } } =x$ , $\frac { x }{ x-2013 } =x$

Solving for $x$ gives $x=2014\quad or\quad 0$

Since $x>0$ , $x=\boxed { 2014 }$

By the way, can anyone tell me what this method is called?