We have the known result (see e.g. Abramowitz and Stegun)
$$P_{2n}(0)=\left(-\frac14\right)^n\binom{2n}{n}$$
Substituting this result into your sum (while also exploiting the oddness of the odd-order Legendre polynomials) yields
$$\sum_{k=0}^\infty\frac{2k+1}{2k+2}\binom{2k}{k}^2\left(\frac1{16}\right)^k$$
which Mathematica says is divergent, and that is certainly because the central binomial coefficients grow quite quickly ($\binom{2k}{k}\sim\frac{4^k}{\sqrt{\pi k}}$).
Nevertheless, we can ask Mathematica to evaluate the sum
$$\sum_{k=0}^\infty\frac{2k+1}{2k+2}\binom{2k}{k}^2 z^k$$
Sum[(((2 k + 1) Binomial[2 k, k]^2)/(2 k + 2)) x^k, {k, 0, ∞}] // FullSimplify
(EllipticK[16 x] - EllipticE[16 x])/(8 π x)
where we have obtained a result in terms of complete elliptic integrals.
Unfortunately, it is also known that $K(m)$ exhibits a logarithmic singularity at $m=1$, so we again reach the conclusion that the original sum is indeed divergent.
SumConvergence[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], i, Assumptions -> Element[i, Integers] && Mod[i, 2] == 0]! – PlatoManiac Aug 28 '13 at 18:17NSum[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], {i, 0, Infinity}, NSumTerms -> 1000]. Similar issue is discussed also here Precision differences – Artes Aug 28 '13 at 19:48exp = Sum[(i + 1)/(i + 2) LegendreP[i, 0] LegendreP[i, 0], {i, 0, k}]; exp /. k -> 30000 // Nwith the result your code gives! – PlatoManiac Aug 28 '13 at 20:14