Proving a Legendre function using generating function 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?
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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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}
