Tackle this series I am looking for the exact value or a smart approximation(if you have a good idea) of the following series:
$$\sum_{n=0}^\infty \frac{1}{2n+1} (P_{n+1}(0)-P_{n-1}(0))$$ where $P_n$ is the n-th Legendre polynomial and $P_{-1}(0)=-1$
 A: Using a CAS, write
$$ P_n(0) = \frac{\sqrt{\pi}}{\Gamma(\frac32+\frac n2)\Gamma(-\frac n2)}, $$
so that
$$ P_{n+1}(0) - P_{n-1}(0) = \frac{(2n+1)\sqrt\pi}{n\Gamma(-\frac n2)\Gamma(\frac32+\frac n2)}. $$
Here $P_{-1}(0) = 1$ by convention.
The term $n=0$ is $-1$, all terms with even $n$ vanish because of the poles of the gamma functions, and the sum over the odd terms is
$$\sum_{k\geq0} \frac{-\sqrt\pi}{2\Gamma(\frac12-k)\Gamma(2+k)} = 1 - \sqrt{2}. $$
Finally, I need to add $2$ because $P_{-1}(0)=1$ above and $P_{-1}(0)=-1$ in the question. Hence the sum is
$$2-\sqrt{2}. $$
A: Let us use the relation
$$\left(\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}\right)'=P_n(x)$$
and consider the series 
$$S(x)=\sum_{n=1}^{\infty}\frac{P_{n+1}(x)-P_{n-1}(x)}{2n+1}.$$
(Note that since $\forall n\geq 0$ one has $P_n(1)=1$, we have $S(1)=0$). Now, using the generating function $\displaystyle \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^{\infty}P_n(x)t^n$ of Legendre polynomials, we find
$$S'(x)=\sum_{n=1}^{\infty}P_n(x)=\frac{1}{\sqrt{2-2x}}-1.$$
Therefore
$$S(0)=-\int_0^1\left(\frac{1}{\sqrt{2-2x}}-1\right)dx=1-\sqrt{2}.$$
But the sum we want to compute is nothing but
$$\sum=\left(P_1(0)-P_{-1}(0)\right)+S(0)=2-\sqrt{2}.$$
A: Since $P_{n+1}(0)=-\frac{n}{n+1}P_{n-1}(0)$ from the recurrence relation, you want
$$
\begin{align*}
\sum_{n=0}^\infty -\frac{\frac{n}{n+1}+1}{2n+1}P_{n-1}(0)
&= \sum_{n=0}^\infty -\frac{P_{n-1}(0)}{n+1}\\
&= \sum_{n=0}^\infty (-1)^2\frac{1}{n+1}\frac{n-2}{n-1}P_{n-3}(0)\\
&= \sum_{n=0}^\infty (-1)^{k+1}\frac{(n-2)(n-4)\dots(n-2k)}{(n+1)(n-1)\dots(n+1-2k)}P_{n-2k}(0)
\end{align*} 
$$
From there, based on $P_0(0)$ and $P_{-1}(0)$, you should be able to find a closed form expression (provided i didn't mess up with the derivation above).
