Legendre polynomial with argument zero $P_n ( 0 )$ I want to find the expression for $P_n(x)$ with $x = 0$, ie $P_n(0)$ for any $n$. The first few non-zero legendre polynomials with $x=0$ are 
$P_0(0) = 1$, $P_2(0) = -\frac{1}{2}$, $P_4(0) = \frac{3}{8}$, $P_6(0) = -\frac{5}{16}$, $P_8(0) = \frac{35}{128}$ but I can't find a relationship between them to write as an equation for arbitrary $n$. Any help is appreciated.


One way to define Legendre polynomials is through its generating function:
$$\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^\infty P_n(x)t^n$$
Together with following sort of well known expansion:
$\displaystyle\;\frac{1}{\sqrt{1-4z}} = \sum_{k=0}^\infty \binom{2k}{k} z^k$,
we have
$$\sum_{n=0}^\infty P_n(0) t^n = \frac{1}{\sqrt{1+t^2}} = \sum_{k=0}^\infty \binom{2k}{k}\left(-\frac{t^2}{4}\right)^k
\quad\implies\quad
P_n(0) = \begin{cases}
\frac{(-1)^k}{4^k} \binom{2k}{k}, & n = 2k\\
0, & \text{ otherwise }
\end{cases}$$
