An identity related to Legendre polynomials Let $m$ be a positive integer. I believe the the following identity
 $$1+\sum_{k=1}^m (-1)^k\frac{P(k,m)}{(2k)!}=(-1)^m\frac{2^{2m}(m!)^2}{(2m)!}$$
 where $P(k,m)=\prod_{i=0}^{k-1} (2m-2i)(2m+2i+1)$, is true, but I don't see a quick proof. Anyone?
 A: Clearly $P(k,m) = (-1)^k 4^k \cdot (-m)_k \cdot \left(m+\frac{1}{2}\right)_k$, where $(a)_k$ stands for the Pochhammer's symbol. Thus the sums of the left-hand-side of your equality is
$$
 1 + \sum_{k=1}^\infty (-1)^k \frac{P(k,m)}{(2k)!} = \sum_{k=0}^\infty 4^k \frac{(-m)_k \left(m+\frac{1}{2}\right)_k}{ (2k)!} = \sum_{k=0}^\infty \frac{(-m)_k \left(m+\frac{1}{2}\right)_k}{ \left(\frac{1}{2}\right)_k k!} = {}_2F_1\left( -m, m + \frac{1}{2}; \frac{1}{2}; 1 \right) = 
$$
Per identity 07.23.03.0003.01 ${}_2F_1(-m,b;c;1) = \frac{(c-b)_m}{(c)_m}$:
$$
  {}_2F_1\left( -m, m + \frac{1}{2}; \frac{1}{2}; 1 \right) = \frac{(-m)_m}{\left(\frac{1}{2}\right)_m} = \frac{(-1)^m m!}{ \frac{(2m)!}{m! 2^{2m}} } = (-1)^m \frac{4^m}{\binom{2m}{m}}
$$
The quoted identity follows as a solution of the contiguity relation for $f_m(z) = {}_2F_1(-m,b;c;z)$:
$$
    (m+1)(z-1) f_m(z) + (2+c+2m-z(b+m+1)) f_{m+1}(z) - (m+c+1) f_{m+2}(z) = 0
$$
Setting $z=1$ and assuming that $f_m(1)$ is finite the recurrence relation drops the order, and can be solved in terms of Pochhammer's symbols.
