About binomial identity

Question

During my research, I checked using Maple that we have numerically for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$

$$ \displaystyle \frac{1}{2} \displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} C_{n+p+t}^{n} \displaystyle \frac{(-1)^{p}}{p}= \displaystyle C_{n+t}^n\displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}$$

I think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?

Is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C_m^k=\displaystyle \frac{m!}{k!(m-k)!}$

Answer 1

We have $$C_{n}^{p+t}C_{n+p+t}^n=[y^{n-t}x^tz^t](x+y+z)^{n+t}\cdot \left(\frac{(x+y+z)y}{xz}\right)^p,$$ where $[M]f$ is a coefficient of monomial $M$ in the polynomial or series $f$. Multiplying by $(-1)^p/p$ and summing by all $p=1,2,\ldots$ we get $$\sum_{p=1}^\infty (-1)^p p^{-1}C_{n}^{p+t}C_{n+p+t}^n=[y^{n-t}x^tz^t] (x+y+z)^{n+t}\log\left(1+\frac{(x+y+z)y}{xz}\right). $$ It remains to use the identity $$ \log\left(1+\frac{(x+y+z)y}{xz}\right)=\log\left(1+\frac{y}x\right)+\log\left(1+\frac{y}z\right) $$ to get twice the right hand side of your formula: $$ [y^{n-t}x^tz^t](x+y+z)^{n+t}\log\left(1+\frac{y}x\right)=C_{n+t}^t [y^{n-t}x^t](x+y)^{n}\log\left(1+\frac{y}x\right), $$ and $$ [y^{n-t}x^t](x+y)^{n}\log\left(1+\frac{y}x\right)=\sum_{p=1}^{n-t} \frac{(-1)^p}p [y^{n-t-p}x^{t+p}] (x+y)^{n}= \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p} $$

We should explain what is the ring the series above belong to, let it be the ring of power series in $x,z$ and $y/(xz)$.