A calculus problem by Peter Orton

$x^0 + x^1 + x^2 + x^3 + x^4 + ....... = 2015$

$=> 1 + x + x^2 + x^3 + x^4 + ....... = 2015$

$=> x + x^2 + x^3 + x^4 + ....... = 2014$

$=> x(1+x + x^2 + x^3 + x^4 + ....... ) = 2014$

But, $1 + x + x^2 + x^3 + x^4 + ....... = 2015$

So, $2015x = 2014 => x = \frac{2014}{2015}$

$x = \frac{2014}{2014+1} => a = 2014$

$2015={ x }^{ 0 }+{ x }^{ 1 }+{ x }^{ 2 }+{ x }^{ 3 }\dots=\\ 2015=1+x+{ x }^{ 2 }+{ x }^{ 3 }\dots$

and

$\frac { 2015 }{ x } =\frac { 1 }{ x } +x+{ x }^{ 2 }+{ x }^{ 3 }\dots$

On subtracting the two equations,

$2015-\frac { 2015 }{ x } =-\frac { 1 }{ x } +1-1+x-x+{ x }^{ 2 }-{ x }^{ 2 }+{ x }^{ 3 }-{ x }^{ 3 }\dots =\\ 2015-\frac { 2015 }{ x } =-\frac { 1 }{ x } \\ 2015x-2015=-1\\ 2015x = 2014\\ x=\frac { 2014 }{ 2015 }$

Given,

$x=\frac { a }{ a+1 } =\\ x=\frac { 2014 }{ 2014+1 } \\ \therefore a=2014$