Proof regarding power series  Assume that $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is convergent in $(-1,1)$ and that $f(1/k) = 0$ for all $k\in \mathbb{N}$. Show that $c_n = 0$ for all $n\geq 0$.
I'm not sure what I can use to prove this. I've tried creating various alternating series, none of which sum to 0, but I'm sure there's a theorem of some sort that would be more useful than attempting to prove this by hand.
 A: Hint: Use mean value theorem (to be more precisely, it should be Rolle's Theorem) to show that there exists a sequence $a_n \rightarrow 0$ such that $f'(a_n) = 0$. 
Inductively, you could prove $f^{(n)}(0) = 0$ for all $n\geq 0$. 
A: Let $f$ be represented by the series
$$f(x)=\sum_{n=0}^\infty c_nx^n$$
for $|x|<1$.  Furthermore, we are given that $f(1/k)=0$ for all $k\in \mathbb{N}$.  Clearly we have
$$0=f(1/k)=\sum_{n=0}^\infty c_n\left(\frac1k\right)^n\tag 1$$

Letting $k\to \infty$ in $(1)$ reveals that $c_0=0$.   Then, we have
$$f(1/k)=\sum_{n=1}^\infty c_n\left(\frac1k\right)^n \tag 2$$
whereupon multiplying both sides of $(2)$ by $k$ yields
$$kf(1/k)=\sum_{n=1}^\infty c_n\left(\frac1k\right)^{n-1}\tag 3$$
Noting that the left-hand side of $(3)$ is $0$, we take the limit as $k\to \infty$ of the right-hand side of $(3)$ to find that $c_1=0$.

Continuing recursively, we have for all $m\ge 1$
$$k^mf(1/k)=\sum_{n=m}^\infty c_n\left(\frac1k\right)^{n-m}\tag 4$$
whence letting $k\to \infty$ reveals that $C_m=0$.
And we are done!
