I met the following infinite summation and Matheamtica is able to simplify it as(Mathematica's notation for convenience): $$\sum _{j=0}^{\infty } \frac{\text{Pochhammer}[1-\gamma ,j]}{j!}\frac{\alpha \gamma }{(t+\alpha +j \alpha )^2}=-\frac{\gamma \text{Gamma}\left[\frac{t+2 \alpha }{\alpha }\right] \text{Gamma}[\gamma ] \left(\text{PolyGamma}\left[0,1+\frac{t}{\alpha }\right]-\text{PolyGamma}\left[0,1+\frac{t}{\alpha }+\gamma \right]\right)}{(t+\alpha ) \text{Gamma}\left[\frac{t+\alpha +\alpha \gamma }{\alpha }\right]},$$
The polygamma function is defined as $$\text{PolyGamma}[0,z+1]=-\gamma +\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{z+n}\right).$$
Any tips on how to see this? Thanks in advance.
Deciphering the Mathematica's notation (not at all convenient to read I think), $$\text{Pochhammer}[1-\gamma,j]:=(1-\gamma)\ldots(j-\gamma)=(-1)^jj!\binom{\gamma-1}{j},\\\text{PolyGamma}[0,z+1]:=\psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)},$$ your sum is equal to $(\gamma/\alpha)S(\gamma,1+t/\alpha)$, where, for $a,b>0$, $$S(a,b)=\sum_{j=0}^{\infty}\binom{a-1}{j}\frac{(-1)^j}{(b+j)^2}=\sum_{j=0}^{\infty}(-1)^j\binom{a-1}{j}\int_0^1 t^{b+j-1}\ln\frac{1}{t}\,dt\\=\int_0^1 t^{b-1}\ln\frac{1}{t}\sum_{j=0}^{\infty}\binom{a-1}{j}(-t)^j\,dt=\int_0^1 t^{b-1}(1-t)^{a-1}\ln\frac{1}{t}\,dt\\=-\frac{\partial}{\partial b}\mathrm{B}(a,b)=\mathrm{B}(a,b)\big(\psi(a+b)-\psi(b)\big),$$ where $\mathrm{B}(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$ is the beta function.