Show that $zF(z)=F(z+1)$ for $z\in \mathbb{C}\setminus\{0,-1,\dots\}$. Here is my attempt: \begin{equation} \sum_{n=0}^\infty \frac{(-1)^n}{n!}\cdot \frac{z}{z+n}+z\int_1^\infty t^{z-1}e^{-t}\ dt \end{equation} Let us look at the second expression first: \begin{equation} z\int_1^\infty t^{z-1}e^{-t}=[t^z\cdot e^{-t}]_1^\infty+\int_1^\infty t^ze^{-t}=e^{-1}+\int_1^\infty t^ze^{-t} \end{equation} I, however, can't simplify the first expression to equal $\sum_{n=0}^\infty \frac{(-1)^n}{n!}\cdot \frac{1}{z+1+n}-e^{-1}$.
Any help would be greatly appreciated.
Here is my approach. Let consider the RHS: $$I: = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\cdot \frac{1}{z+1+n}+e^{-1} =- \sum_{n=1}^\infty \frac{(-1)^{n}}{n!}\cdot \frac{n}{z+n}+e^{-1} $$ $$=\sum_{n=1}^\infty \frac{(-1)^{n}}{n!}\frac{z}{z+n}-\sum_{n=1}^\infty \frac{(-1)^{n}}{n!} +e^{-1}$$ Applying the Maclaurin series for $e^{-1}$, we obtain: $$I = \sum_{n=1}^\infty \frac{(-1)^{n}}{n!}\frac{z}{z+n}+1=z \sum_{n=0}^\infty \frac{(-1)^{n}}{n!}\frac{1}{z+n} $$ Since: $$\int_1^\infty t^{z}e^{-t}\ dt = - t^{z}e^{-t}\Bigg|_1^\infty + z\int_1^\infty t^{z-1}e^{-t}\ dt $$ $$= e^{-1} + z\int_1^\infty t^{z-1}e^{-t}\ d$$ Hence, adding terms by terms will yield desired equality.