Show that on any bounded subset $K\subseteq \Bbb{C}$, the sequence is a Cauchy with respect to the norm $\|\cdot\|_{\infty,K}$ Define $S_N (z)=\int_{1}^{N}t^z e^{-t}dt$ and $f(z)=\int_{1}^{\infty} t^z e^{-t}dt$. 
Show that on any bounded subset $K\subseteq \Bbb{C}$, we have $S_N$ is a Cauchy sequence with respect to the norm $\|\cdot\|_{\infty,K}$. In other words, ${S_N}$ satisfies the hypothesis of the "Cauchy Criteria" so that $S_N \rightarrow f$ uniformly on $K$, and in particular $f(z)$ is well defined on all $z\in \Bbb{C}$
Since $K$ is bounded, we can assume that $|\textrm{Re }z|\leq L$ for some $L>0$ and for all $x\in K$. But I don't know how to proceed further from here.
 A: Take $N \in \mathbb{N}$ large, and let $M\in \mathbb{N}$ be any. Let also $K\subset \mathbb{C}$ be a bounded set, in particular (as you mentioned in your post) we have
$$
|\mathrm{Re} z| \leq L \ \text{ for any } z\in K \tag{1}
$$
where $L>0$ is a fixed constant depending on $K$. We have
$$
|S_{N+M}(z) - S_N(z)| =  \left| \int\limits_N^{N+M} t^z e^{-t} dt \right| \leq \int\limits_{N}^{N+M} |e^{z\ln t}| e^{-t} dt. \tag{2}
$$
Writing $z = x+iy$ with $x,y\in \mathbb{R}$ we get $|e^{z \ln t}| = e^{x \ln t} \leq e^{L \ln t}$ in view of $(1)$. 
Using this bound in $(2)$ we get
$$
|S_{N+M}(z) - S_N(z)| \leq \int\limits_{N}^{N+M} e^{-t +  L \ln t} dt \leq \int\limits_{N}^{N+M} e^{-\frac 12 t} dt = 2 e^{-\frac 12 N} (1- e^{-\frac 12 M}) \tag{3},
$$
where in the second inequality of $(3)$ we used the fact that $-t + L \ln t \leq -\frac 12 t$ when $t>N$ for $N$ large enough. The right-hand side of $(3)$ is independent of $z\in K$ and hence $\{S_N(z)\}$ is Cauchy with respect to supremum norm on $K$, and hence converges to $f$ uniformly on $K$ by the definition of $f$.
