= Solution
The <generating function> gives $P_n(-t)=(-1)^nP_n(t)$, so the odd function $f$ is orthogonal to every even-degree <Legendre polynomial>. For $n\geq1$, integrate the self-adjoint form of the <Legendre differential equation> from zero to one:
$$
[(1-t^2)P_n'(t)]_0^1+n(n+1)\int_0^1P_n(t)dt=0.
$$
Thus $\int_0^1P_n=P_n'(0)/(n(n+1))$. Differentiating the <generating function> in $t$ and then setting $t=0$ gives
$$
\sum_nP_n'(0)z^n=z(1+z^2)^{-3/2},\qquad P_{2m+1}'(0)=(-1)^m\frac{(2m+1)!}{2^{2m}(m!)^2}.
$$
Combining these identities with the normalization of $p_n$ yields
$$
\boxed{\langle f,p_{2m}\rangle=0,\qquad\langle f,p_{2m+1}\rangle=(-1)^m\sqrt{\frac{4m+3}{2}}\frac{(2m)!}{2^{2m}m!(m+1)!}.}
$$
For the <Legendre expansion of the sign function>, $\|f\|_2^2=2$, so the <Parseval identity> gives
$$
2=\sum_{m\geq0}\frac{4m+3}{2}\frac{((2m)!)^2}{2^{4m}(m!(m+1)!)^2}.
$$
Therefore the requested sum equals
$$
\boxed{4.}
$$
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