Proving the closed form for an infinite sum (related to Chebyshev polynomials)

Question

How do I prove the following identity? For $y\not= 0$, we have $$ \sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right) = \dfrac{1}{(x+y-1)(x-y-1)}. $$

I am trying to find the closed form for the left hand side. Thanks in advance!

Answer 1

$$ \begin{align} \sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right) &= \sum_{n=0}^{\infty} \sum_{k=0}^n (x+y)^{k}(x-y)^{n-k} \\ &= \sum_{k=0}^\infty \sum_{n=k}^{\infty} (x+y)^{k}(x-y)^{n-k}\\ &= \sum_{k=0}^\infty (x+y)^{k} \sum_{n=k}^{\infty} (x-y)^{n-k}\\ &= \sum_{k=0}^\infty (x+y)^{k} \sum_{m=0}^{\infty} (x-y)^{m}\\ &= \frac{1}{1-(x-y)}\sum_{k=0}^\infty (x+y)^{k} \\ &= \frac{1}{1-(x-y)}\cdot\frac{1}{1-(x+y)}\\ &= \frac{1}{(x+y-1)(x-y-1)} \end{align} $$