<s>[INST] Proving the closed form for an infinite sum (related to Chebyshev polynomials) 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! 
 [/INST] $$
\begin{align}
\sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right) 
&= \sum_{n=0}^{\infty} \sum_{k=0}^n (x+y)^{k}(x-y)^{n-k} \\
&= \sum_{k=0}^\infty \sum_{n=k}^{\infty} (x+y)^{k}(x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{n=k}^{\infty} (x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{m=0}^{\infty} (x-y)^{m}\\
&= \frac{1}{1-(x-y)}\sum_{k=0}^\infty (x+y)^{k} \\
&= \frac{1}{1-(x-y)}\cdot\frac{1}{1-(x+y)}\\
&= \frac{1}{(x+y-1)(x-y-1)} 
\end{align}
$$
</s>