Prove $\sum \frac{\cos nz}{n!}$ converges on compact sets

Question

Prove that

$$\sum_{n=1}^{\infty} \displaystyle\frac{\cos nz}{n!}$$

Is an entire function, which means that it uniformly convergent on compact sets

Answer 1

Assuming $z\in\mathbb{R}$, your series is the real part of $$ \sum_{n\geq 0}\frac{e^{inz}}{n!} = e^{e^{iz}} = e^{\cos(z)}\left(\cos(\sin z)+i\sin(\sin z)\right),$$ so: $$ \sum_{n\geq 0}\frac{\cos(nz)}{n!} = e^{\cos(z)}\cos(\sin(z)) $$ holds for any $z\in\mathbb{C}$ by analytic continuation (or simply by writing $\cos(nz)$ as $\frac{e^{iz}+e^{-iz}}{2}$), and the RHS is an entire function, since the product or composition of two entire functions is still an entire function. At last, if we have an entire function, its Taylor series (centered at any $z_0\in\mathbb{C}$) is uniformly convergent to the function over any compact subset of $\mathbb{C}$. Here $\|z\|\leq R$ grants $$ \|\cos(nz)\|\leq 2 e^{nR} $$ and $$\sum_{n\geq 0}\frac{ 2 e^{nR}}{n!} = 2 e^{e^R} < +\infty. $$

Answer 2

Sketch:

You can use that $\cos(nx + n iy) = \cos nx \cosh ny - i\sin nx \sinh ny$ to obtain that $|\cos nz| \leq e^{|ny|}$. Since factorials grow faster than exponentials, the complex version of the Weierstrass $M$-test will imply the result.