An infinite function is asked an uninfinite question that may hold an infinite answer

This solution has two parts:

PART 1: REWRITING $f(x)$ TO A SHORTER FORM

$f(x)$ can be rewitten as: $f(x)= \left ( \frac { 1 }{ x } \right) +\left ( \cfrac { 1 }{ x } \right) ^{ 2 }+\left ( \cfrac { 1 }{ x } \right) ^{ 3 }+\dots$ Notice that $f(x)$ is actually the sum of an infinite geometric sequence (based on the above equation) with a first term and common ratio $\left ( \frac { 1 }{ x } \right )$

By using the formula for the sum of an infinite geometric sequence, $S_{\infty}= \frac { first\space term }{1 - common\space ratio }, f(x)$ can be expressed as: $f(x)=\frac { \left( \cfrac { 1 }{ x } \right) }{ 1-\left( \cfrac { 1 }{ x } \right) }$ Which can be simplified to become: $f(x)=\cfrac { 1 }{ x-1 }$

PART 2: GETTING THE INVERSE OF $f(x)$

To get the inverse of a function, $f(x)$ will be substituted by $x$ and $x$ will be replaced by $f^{-1}(x)$ Then, we have: $x=\frac { 1 }{ f^{ -1 }(x)-1 }$ Manipulating this equation gives: ${ f }^{ -1 }(x)=\cfrac { 1+x }{ x }$ Therefore, ${ f }^{ -1 }(2015)=\cfrac { 1+2015 }{ 2015 } =\boxed { \cfrac { 2016 }{ 2015 } }$

Its an infinite g.p. Use the formula for infinite g.p And you will get the answer x = 2016/2015