I must prove that $\int_{-1}^1 (1-2xt+t^2)^{-1/2}P_n(x)dx=\frac{2t^n}{2n+1}$.
I know that the generating function is $(1-2xt+t^2)^{-1/2}=\sum_{n=0}^\infty P_n(x)t^n$. I also know that the orthogonality property when l=m gives $\frac{2}{2n+1}$.
I get to where I have $\sum_{n=0}^\infty t^n(\frac{2}{2n+1})$.
Any ideas on how to get the correct answer or where I may be going wrong?
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You are calculating for 'a given' $\ds{\,\mathrm{P}_{n}\pars{x}}$ $\ds{~\pars{\mbox{namely, for 'a given'}\ n}~}$. You must use another index for the generating function $\ds{~\pars{n\ \mbox{is a constant}}~}$ such as $\ds{m}$ which I'll use below !!!.
\begin{align} &\color{#f00}{% \int_{-1}^{1}\pars{1 - 2xt + t^{2}}^{-1/2}\,\,\,\mathrm{P}_{n}\pars{x}\,\dd x} = \int_{-1}^{1}\overbrace{% \bracks{\sum_{m = 0}^{\infty}t^{m}\,\mathrm{P}_{m}\pars{x}}} ^{\ds{\pars{1 - 2xt + t^{2}}^{-1/2}}} \,\mathrm{P}_{n}\pars{x}\,\dd x \\[5mm] = &\ \sum_{m = 0}^{\infty}t^{m}\ \underbrace{% \int_{-1}^{1}\,\mathrm{P}_{m}\pars{x}\,\mathrm{P}_{n}\pars{x}\,\dd x} _{\ds{2\,\delta_{mn} \over 2n + 1}}\ =\ \color{#f00}{{2\,t^{n} \over 2n + 1}} \end{align}
