2019
Problem - 4530
Compute the limit of the power series below as a rational function in $x$:
$$1\cdot 2 + (2\cdot 3)x + (3\cdot 4)x^2 + (4\cdot 5)x^3 + (5\cdot 6)x^4+\cdots,\qquad (|x| < 1)$$
Because $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots $$
is convergent, therefore it is possible to take derivative on both sides. This gives
$$\frac{1}{(1-x)^2}=1+2x + 3x^2 + 4x^3 + \cdots$$
Taking derivative on both sides again yields:
$$\frac{2}{(1-x)^3}=2 + (2\cdot 3)x + (3\cdot 4)x^2 + (4\cdot 5)x^3 + \cdots$$
Therefore, the answer is $\boxed{\frac{2}{(1-x)^3}}$.