Explaining why there is no power series $f(z)=\sum^{\infty}_{n=0}C_{n}z^{n}$

Question

As the title suggests, I would like to understand why there is no power series $f(z)=\sum^{\infty}_{n=0}C_{n}z^{n}$ such that $f\big(\frac{1}{k}\big)=1$ $\forall$ $k=2,3,4,...$ and $f'(0)>0$?

A solution to this problem was given but I am having a difficult time trying to understand it.

$Solution$: The centre of the power series is at $0$. Since $$\lim_{k\to \infty}\frac{1}{k}=0, (*)$$
$f(z)=1+0z+0z^{2}+0z^{3}+ \dots (**)$

Then, $f'(z)=0$.

So, $f'(0) \not >0.$ (***).

May I ask:
1. Why do we need to find the limit of $\frac{1}{k}$ as $k \to \infty$? Why does (*) imply that $f(z)=1+0z+0z^{2}+0z^{3}+ \dots$?
2. Consequently, how do we know that the constant term of (**) is $1$?
3.What is the value of $f(0)$? Why is it $\not >0?$

Any help would be appreciated!

Answer 1

A such power series is continuous. So, if $f(1/k)=1$ for all $k$, we must have $f(0)=1$. Now, considering $g(z)=f(z)-1$, we have an accumulation of zeroes for $g(z)$ this forces your series to be constant (zeroes and poles are isolated) $g\equiv 0$ and then $f\equiv 1$, so $f'(0)=0$.

Do not hesitate to interact if something is not clear for you.

Answer 2

(Assuming you do not know the identity theorem)

If the power series converges in a neighbourhood of $0$, it converges absolutely and to a continuous function in such a neighbourhood. Consequently, $C_0=f(0)=\lim_{k\to\infty}f(\frac1k)=1$.

Additionally, we can differentiate termwise (again because of absolute convergence), i.e., $f$ is differentiable near $0$ and $f'(z)=\sum_{n=0}^\infty (n+1)C_{n+1}z^n$. By Rolle, $f'(\xi_k)=0$ for some $\xi_n$ between $\frac1k$ and $\frac1{k+1}$, hence $C_1=f'(0)=\lim_{k\to\infty}f'(\xi_k)=0$.

The above argument can be repeated to conclude $C_n=0$ for all $n>0$.