I have this sum $$\left(N+1\right)^{2}\underset{j=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{j}}{2j+1}\dbinom{N}{j}\dbinom{N+j}{j-1}\underset{i=1}{\overset{N}{\sum}}\frac{\left(-1\right)^{i}}{\left(2i+1\right)\left(i+j\right)}\dbinom{N}{i}\dbinom{N+i}{i-1}$$ and numerical test indicates that is equal to $$\frac{\left(N+1\right)N}{\left(2N+1\right)^{2}}$$ but I'm not able to prove it. Thank you

  • $\begingroup$ Where did this come from? It may help someone else answer it. $\endgroup$ – snulty Jul 17 '14 at 18:01

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