Suppose that there are entire functions $\{f_n\}$ so that for all complex numbers $x+iy$ $$\sum_{n=1}^{\infty} |f_n(x+iy)|^{\frac{1}{n}} \leq e^x$$ Show that $f(z)=\sum_{n=1}^{\infty} f_n(x+iy)$ is analytic on $\{\Re(z) < 0\}$ and has period $2\pi i$.
I don't know how to grid rid of $\frac{1}{n}$. Can anybody give me some ideas? How to prove it's uniformly convergent on any compact subset? And how to prove it's periodic?
For every $n$, the function $e^{-nz}f_n(z)$ is bounded by $1$ in $\mathbb C$, hence constant by Liouville's theorem. In other words, $f_n(z)=c_n e^{nz}$ where $|c_n|\le 1$.
Any compact subset $K$ of the left halfplane is contained in some halfplane $x\le x_0$ with $x_0<0$. On $K$ we have $|f_n(z)|\le (e^{x_0})^n$ for all $n$. Since $e^{x_0}<1$, the Weierstrass test for uniform convergence applies. And since all $f_n$ are $2\pi i$-periodic, so is $f$.