∑k=1n+1(n+1)!⁢(k-1)!(n-k+1)!⁢(-a-n)k⁢zn-k+1-∑k=0-a-n-1(a+n+1)k(n+2)k⁢k!⁢zn+k+1⁢(ln⁡z+ψ⁡(-a-n-k)-ψ⁡(1+k)-ψ⁡(n+k+2))+(-1)n-a⁢(-a-n-1)!⁢∑k=-a-n∞(k+a+n)!(n+2)k⁢k!⁢zn+k+1,