Αn entire function as an infinite sum of entire functions

Question

Please help me to attack this problem from Complex Analysis:

Let be $f$ be an entire function, such that $f(0)=0$. Prove that the series: $$\sum_{n=1}^{\infty}f\left(\frac{z}{n^2}\right)$$ defines an entire function.

Answer 1

Hint: show that the series converges uniformly on the disk $|z| \le R$ for any $R$.

Answer 2

Let $$ f(z)=a_1z+a_2z^2+\cdots+a_nz^n+\cdots $$ and set $$ g_n(z)=\sum_{k=1}^n f\left(\frac{z}{n^2}\right). $$ By virtue of the integral test for series $$ \sum_{k=m}^n \frac{1}{n^{2j}}<\sum_{k=m}^\infty \frac{1}{k^{2j}}\le \int_{m-1}^\infty\frac{dx}{x^{2j}}=\frac{1}{(2j-1)(m-1)^{2j-1}}\le \frac{1}{m-1}. $$

Let $R>0$ arbitrary, then for $|z|\le R$ and $\,n\ge m>1$, we have $$ |g_m(z)-g_n(z)|\le \sum_{k=m}^n\left|\,f\left(\frac{z}{k^2}\right)\right| \le \sum_{k=m}^n \sum_{j=1}^\infty \frac{|a_j||z|^j}{k^{2j}}\le \frac{1}{m-1} \sum_{j=1}^\infty |a_j|R^j. $$

This implies that the sequence of entire functions $\{g_n(z)\}$ is locally uniformly Cauchy, and hence locally uniformly convergent to an entire function.