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i'm unable to prove the following : $\forall n$ integer $\geq 3$,

$ \displaystyle \displaystyle \sum_{s=1}^n \sum_{j=n-s+1}^n \displaystyle \frac{ (\binom n j )^2 \binom {n+j} n }{s-n+j} ( \displaystyle \sum_{t=1}^n \displaystyle \frac{1}{j+t} +2 \displaystyle \sum_{t=0,t \neq j}^n \displaystyle \frac{1}{t-j}) $ $ =\displaystyle \sum_{s=0}^n \displaystyle \sum_{r=s+1}^n \displaystyle \frac{ \binom n r \binom n s (-1)^{r+s} }{r-s} {\Big{(}} \binom {n+r} n + \binom {n+s} n {\Big{)}} \displaystyle \sum_{t=s+1}^r \frac{1}{t} $

Any help? Is that a new result?

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  • $\begingroup$ Is there a q-analogue of this? $\endgroup$
    – Per Alexandersson
    Mar 11, 2019 at 9:47
  • $\begingroup$ i don't know (i have found this ) making computation $\endgroup$
    – mamiladi
    Mar 11, 2019 at 13:55
  • $\begingroup$ $ \displaystyle \sum_{s=0}^n \binom n s (-1)^s \binom {s+n} n \sum_{t=0}^{n-1} \displaystyle \frac{1}{t+s+1}=0$ $\endgroup$
    – mamiladi
    Mar 11, 2019 at 14:06
  • $\begingroup$ for all value $ 0 \leq t \leq n-1$ ( the explain of that is simple we can use orthonalite legendre polynom $P_n to x^t$ and expand term inside integral $ P_n*x^t $ and integrate term by term) $\endgroup$
    – mamiladi
    Mar 11, 2019 at 14:10
  • $\begingroup$ this may be can usefll at the second member to proof the identity above $\endgroup$
    – mamiladi
    Mar 11, 2019 at 14:11

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