I understand the function is:
#f(x) = (x/(2-x))^3#
we can write it as:
#f(x) = (x(1/(2-x)))^3 =x^3 (1/(2-x))^3#
Consider now a primitive of #g_0(x) = (1/(2-x))^3#:
#g_1(x) = int_(-oo)^x (1/(2-t))^3 dt = - int_(-oo)^x (1/(2-t))^3 d(2-t) = 1/2[1/(2-t)^2]_(-oo)^x = 1/2 1/(2-x)^2#
and again:
#g_2(x) = int_(-oo)^x (1/(2-t))^2 dt = - int_(-oo)^x (1/(2-t))^2 d(2-t) = [1/(2-t)]_(-oo)^x = 1/2 1/(2-x)#
Now #g_2# can be expressed as the sum of a geometric series:
#g_2(x) = 1/2 1/(2-x) = 1/4 1/(1-x/2) = 1/4 sum_(k=0)^oo (x/2)^k = sum_(k=0)^oo x^k/2^(k+2)#
converging for #absx < 2#. Inside the interval of convergence we can differentiate the series term by term and obtain a series with the same radius of convergence, so that:
#g_1(x) = d/dx (sum_(k=0)^oo x^k/2^(k+2)) = sum_(k=1)^oo (kx^(k-1))/2^(k+2)# with #abs x < 2#
and differentiating again:
#g_0(x) = sum_(k=2)^oo (k(k-1)x^(k-2))/2^(k+2) # with #abs x < 2#
Now:
#f(x) = x^3 g_0(x) = x^3 sum_(k=2)^oo (k(k-1)x^(k-2))/2^(k+2) = sum_(k=2)^oo (k(k-1)x^(k+1))/2^(k+2)# with #abs x < 2#
