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?
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$\begingroup$ You don't want that $(-1)^n$: it should be just $-1/(4n^2-1)$. $\endgroup$– Robert IsraelCommented Jul 22, 2014 at 16:24
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$\begingroup$ This bears some resemblance to Catalan numbers. Maybe you can find and tweak a similar relation among Catalan numbers to get a proof. $\endgroup$– The Masked AvengerCommented Jul 22, 2014 at 17:32
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2 Answers
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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 $2\nu$.)
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Somewhat more generally, Maple 18 says that
$$ \sum_{k=1}^n \dfrac{(-1)^k k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} = -{\frac {\Gamma \left( v/2+2 \right) \Gamma \left( n-v/2 \right) }{ \left( v+2 \right) \Gamma \left( 1-v/2 \right) \Gamma \left( 1+v/2+ n \right) }} $$ which for $v=1$ gives you the desired identity.
It appears that Maple is using Zeilberger's algorithm.
answered Jul 22, 2014 at 19:09