Determine Positive Integer in a Series Let $a_{n} = 4 - 3n,$ $\forall$ Integers $n \geq 1.$ 
$$f(x, y) = x + \sum_{i=1}^{\infty}\left [\left(\frac{\prod_{j=1}^{i}a_{j}}{3^i\cdot i!} \right )x^{a_{i+1}}y^{i}\right]$$
$\forall$  Real $x$ and $y.$ How can I find a positive integer $k$ such that $f(19, k) = 20$.
 A: Consider the function $F(x, y) = \sqrt[3]{x^3 + y}.$ If we write the function in the form $(x^3 + y)^{1/3},$ we can use the generalized version of the binomial theorem to expand this. We get
$$(x^3 + y)^{1/3} = \binom{1/3}{0}x + \binom{1/3}{1}x^{-2}y + \binom{1/3}{2}x^{-5}y^2 \cdots$$
Also,
$$\begin{align*} \binom{1/3}{n} &= \frac{(1/3)(1/3 - 1)(1/3 - 2)\cdots(1/3 - n + 1)}{n!} \\ &= \frac{(1/3)(-2/3)(-5/3)\cdots(a_{n}/3)}{n!} \\ &= \frac{\prod_{i=1}^{n}a_{i}}{3^n \cdot n!}. \\ \end{align*}$$
Finally, notice that the exponent on $x$ for the $n$th term in our expanded expression, excluding the first $x$, actually equals $a_{n+1}$, and the corresponding exponent on $y$ is $n.$ Combining all this, we get
$$\binom{1/3}{0}x + \binom{1/3}{1}x^{-2}y + \binom{1/3}{2}x^{-5}y^2 \cdots = x + \sum_{n=1}^{\infty}\left [\left(\frac{\prod_{i=1}^{n}a_{n}}{3^n\cdot n!} \right )x^{a_{n+1}}y^{n}\right] = f(x, y).$$
Thus, $F(x, y) = f(x, y) = \sqrt[3]{x^3 + y}.$ To solve the problem, we need to find the integer $k$ such that $19^3 + k = 20^3.$ This value is $k = 20^3 - 19^3 = 20^2 + 20(19) + 19^2 = 1141.$
We still need to show that this value of $k$ will indeed make the series equal $20$ i.e. if the series converges or diverges. The binomial expansion of $(x+y)^n$, for a real number $n$ that is not a nonnegative integer, will converge if $\left | x \right | > \left | y \right |$ and diverge otherwise. In this case, since $\left | 19^3 \right | = 6859$ is clearly greater than $1141,$ our series will converge, so $f(19, 1141) = 20.$
