For small values of $n$ ($2\leqslant n\leqslant 5$), the coefficients $a\_k = (-1)^k{n\choose k}{n+k\choose k}$ of the shifted Legendre polynomial $\tilde{P}\_n(x)$ satisfy the identity $\displaystyle\sum\_{k=1}^n \frac {ka\_k}{(n+k)(2k+1)} = \frac{(-1)^n}{4n^2-1}.$ Does that identity hold for all $n$? If so, how might one prove it?



Identity involving shifted Legendre coefficients
If we write the right side of Robert's identity as 
$$(-1)^n\frac{(\nu-2)(\nu-4)\cdots(\nu - 2n+2)}{(\nu+2)(\nu+4)\cdots(\nu+2n)},$$
we see that the identity is a partial fraction expansion of a proper rational function. (This is a littler simpler if we replace $\nu$ with $2\nu$.)


