Is the function an entire function?

Question

Is the following an entire function? (Here $z\in \mathbb{C}$) $$\sum_{n=0}^\infty \frac{2^n}{n!}z^{3n}$$

($***$) So, here I first note that the function is a sum of powers of $z$. Now if I show that the sum converges for all $z\in \mathbb{C}$, the problem will be solved, right?

Using ratio test i get $\frac{a_{n+1}}{a_n}=\frac{2z}{n+1}$ which goes to $0$ as $n$ goes to $\infty$, showing that the function converges. The only problem is that is my statement $(***)$ correct?

Thanks in advance.

Answer 1

The problem might be that you are not using the right way to calculate the radius of convergence. First you need to get it right what $a_n$ means: it is always the coefficient of the $n$-th power term and not anything else, so in your case it should be $a_{3n}=2^n/n!$ and the others are all zero. This also means that d'Alembert ratio test isn't immediately applicable here.

However you can always use Cauchy test even if there are many zero coefficients involved, as Cauchy test only cares about the upper limit instead of the limit itself, hence unaffected by zero coefficients in between at all: $$\limsup_{n\to\infty}(a_n)^\frac1n=\limsup_{n\to\infty}(a_{3n})^\frac1{3n}=\cdots=0. $$ which shows that the RoC is infinity.

Answer 2

The function defined by the sum is entire, yes. For this, it suffices to check that the radius of convergence is infinite (which you have done); the result then follows as this implies that the power series is uniformly convergent everywhere.