Give the largest open set where the function $f(z)=\sum_{n=1}^{\infty} ne^{nz}$ is analytic.
Here are the steps:
For the series to converge it is necessary that $ne^{nz} \rightarrow 0$. But $|e^{nz}|=e^{n\Re(z)}$ so it is necessary that $\Re(z)<0$. The series does converge whenever $\Re(z)<0$. Now we have to prove that it is analytic in ${z \in \Bbb C:\Re(z)<0}$. To prove that the series converges uniformly on compact subsets of ${z\in \Bbb C :\Re(z)<0}$. suppose $K⊂ \{z \in \Bbb C:\Re(z)<0\}$ is compact then there exists $\delta >0$ such that $\Re(z)<−\delta$ for all $z \in K$. Did I miss anything? How ca I apply M-test to get uniform convergence on $K$. I appreciate your kind help. Thank you so much!
Following your steps, for a compact set $K \subseteq {z \in \mathbf{C}: Re(z)<0}$, we can find such a $\delta >0$ such that $Re(z)leq-\delta, \forall z \in K$. Now, note that $\frac{n}{e^{n Re(z)}}\leq \frac{n}{e^{n \delta}} \forall z \in K , n\geq 1 $. Take, $M_n= \frac{n}{e^{n \delta}}.$
Let $$S_N=\sum_{n=1}^{N}M_n=\frac{1}{e^{\delta}}+\frac{2}{e^{2\delta}}+\frac{3}{e^{3\delta}}+...+\frac{N}{e^{N\delta}}=\frac{1-Ne^{-N\delta +1}}{1-e^{-\delta}}$$ Then, $lim_{N\rightarrow\infty}=\frac{1-0}{1-e^{-\delta}}$ which is a finite number. Hence by the $M_n$ test the requirment holds.