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Exist a closed form for $$\left(-1\right)^{N}\underset{i=1}{\overset{N}{\sum}}\left(-1\right)^{i}\dbinom{N}{i}\dbinom{N+i}{i-1}\,\frac{1}{2i+1}?$$ I think I've to use in some way the formula of the shifted legendre polynomials $$P_{N}\left(x\right)=\left(-1\right)^{N}\underset{i=0}{\overset{N}{\sum}}\dbinom{N}{i}\dbinom{N+i}{i}\left(-x\right)^{i}$$ but I'm not sure about it. I'm wrong? Thank you

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  • $\begingroup$ Well, have you tried guessing general answer from first few values? $\endgroup$ – Grigory M Jul 15 '14 at 20:44
  • $\begingroup$ Hint : $\displaystyle{n+i\choose i-1}={-n-2\choose i-1}$ $\endgroup$ – Lucian Jun 29 '18 at 19:59
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For $N\ge 1$ we get the fractions 1/3, 2/15, 1/7, 4/45, 1/11, 6/91, 1/15, 8/153 and so on. So the numerators are 1,2,1,4,1,6,1,8, one's interleaved with the even numbers, and for the denominators 3,15,7,45,11,91,15,153, which are the sequence 3,7,11,15, of numbers == 3 (mod 4) interlaced with the values of 15,45,91,153,.. reproduced in https://oeis.org/A014634. So for odd $N$ the values are $1/(2N+1)$, and for even $N$ they are $N/((N+1)(2N+1))$

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