How to prove that $f $ is entire? Prove that for every $z ∈ \mathbb C$, the series $$
\sum_{n=1}^\infty \frac{\sin(z/n)}{n}$$  converge For $z ∈ \mathbb C$, let $$f(z) =
\sum_{n=1}^\infty \frac{\sin(z/n)}{n}.$$  Prove that f is entire.
I know that $\sin z$ is entire  and $\sin(z/n)$ is also entire. I know that if a sequence of holomorphic
functions converges uniformly on compact sets, then the limit is a holomorphic function. But here 
$\frac{\sin(z/n)}{n}$   converges uniformly on compact sets.
I don't know  how to prove that $f(z)$ is entire .
I would be thankful if someone would give me some hints or if u have time give me the proper proof.
Thanks in advance
 A: As you said, if a sequence of holomorphic functions converges uniformly on compact sets, its limit is holomorphic. This said, we only need to prove that the partial sum
$$ S_k(z) = \sum_{n=1}^k \frac{\sin(z/n)}{n} $$
converges uniformly on $B_R(0)$, for all $R>0$. Fix $R>0$. First we shows that the following series converge
$$ \sum_{n=1}^\infty \frac{\|\sin(z/n)\|_{L^\infty(B_R(0)}}{n}. $$
By compactness, denote by $z_n$ the value that realizes the maximum modulus of $\sin(z/n)$ over the ball. The series reads
$$ \sum_{n=1}^\infty \frac{|\sin(z_n/n)|}{n}, \quad z_n\in B_R(0). $$
Since $\lim_{z\to 1} \frac{\sin(z)}{z}=1$, then
$$ \left| \frac{\sin(z_n/n)}{n} \right| \sim \frac{|z_n|}{n^2} \leq \frac{R}{n^2}. $$
This shows that the series converges. In particular, the partial sums $S_n(z)$ are Cauchy wrt $\|\cdot\|_{L^\infty(B_R(0))}$, so that they converge uniformly to an holomorphic function over compact sets.
We are done, indeed
$$ S_\infty (z) = \sum_{n=1}^\infty \frac{\sin(z/n)}{n} $$
is well defined and holomorphic on $B_R(0)$ for all $R>0$, i.e. is entire.
A: Now that you've been given an answer you should prove this:


If $g$ is an entire function and $g(0)=0$ then the series $\sum_1^\infty\frac{g(z/n)}{n}$ converges uniformly on compact sets


