A holomorphic function which takes real values at $ 1/n $ has real coefficients I'm facing the following problem:
Let $ f : U \rightarrow \mathbb{C} $ be holomorphic ($ U $ is a complex domain and $ 0 \in U $). Suppose that for all $ n = 1,2,3 \dots $ it holds that $ f(1/n) \in \mathbb{R} $. Prove that the coefficients $ a_0, a_1, \dots $ such that $ f(z) = \sum_{i=0}^\infty a_iz^i $ are real.
So, obviously, those $ a_i$ are equal to $ f^{(i)}(0) $. It's easy to see that $ a_0 = f(0) $ is real as a limit of real numbers. So is $ a_1 = f'(0) = \lim\limits_{n \rightarrow \infty} \frac{f(1/n) - f(0)}{1/n} $. 
If I knew that $ f'(1/n) $ is real for all $ n $, I could use the same argument to prove that $ f''(0) \in \mathbb{R} $ and go inductively. However, I can't think of a way to figure this out.
I would appreciate some help
 A: Answered in a comment:

the function $g(z)=\overline{f(\bar z)}$ is holomorphic and $g(1/n)=f(1/n)$ for every $n$, hence $f=g$. — user8268

Here $\{1/n\}$ could be any subset of $\mathbb{R}$ with a limit point.
A: Here is a direct approach using induction.
Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be as you stated in the question.
We know that $f$ is continuous at zero, thus $f(1/k) \to f(0)$ as $k\to\infty$. Since each $f(1/k)$ is real, and the reals are a closed subset of the complex plane, $f(0)=a_0$ must be real.
Now consider $g_1(z) = \frac{f(z)-a_0}{z}$. This is again a Holomorphic function at the origin. Moreover $g_1(0) = a_1$. Thus we know that $g_1(1/k) \to g_1(0)=a_1$ as $k\to\infty$. Since each $g_1(1/k)$ is real, again we have $a_1$ is real.
Now assume that for some $N \ge 1$ we have $a_{N} \in \mathbb{R}$. Write $$g_{N+1}(z) = \frac{ f(z) - \sum_{n=0}^{N} a_n z^N }{z^{N+1}}.$$
Since each $a_n$ is real, and $f(1/k)$ is real for each $n \le N$ and $k$ in $\mathbb{N}$, $g_{N+1}(1/k)$ is real for each $k$. Moreover, $g_{N+1}(1/k) \to a_{N+1}$ as $k \to \infty$.
Thus we have proved that each $a_n$ is real by induction.
