$f\in C^{\infty}$ and $|f^{(j)}(x)|\leq M$- show that if $f(1/k)=0 $ for $k\in \mathbb{N}$  so $f=0$ I'd love your help with the following question:
let $f\in C^{\infty}$ and there's $M$ such that $|f^{(j)}(x)|\leq M$ for every $j\in \mathbb{Z}_{+}$  and for all $ x\in [-1,1]$.
I need to prove tht if $f(1/k)=0 $ for every $k\in \mathbb{N}$, so $f=0$.
O.k, so to be honest I tried Lagrange theorem, Taylor series, I tried to get some things together, but Didn't get any good conclusion.
From the continuous of $f$ we get that $f(0)=0$.
Thank you so much guys.
 A: Taylor's theorem states that 
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1}, $$
where $\xi_L$ is between $a$ and $x$.
This will help us finding the function if we take $a=0$. First note that by Lagrange's theorem for $x >0$ we have $f(x)-f(0)=f'(c)x, \ c \in [0,x]$. Therefore, there exists $c_k \in (0,1/k)$ with $f'(c_k)=0$. We can pick a decreasing subsequence of this sequence, and denote it for simplicity with (the same) $(c_k)$. Since $f'(c_k)=0$ and $c_k \to 0$ it follows that $f'(0)=0$. 
Doing the same think with $(c_k)$ instead of $1/k$ we get that $f''(0)=0$ and inductively $f^{(n)}=0$, for all positive integers $n$.
Pick $x \in [-1,1]$. By Taylor's formula around zero we get that $f(x)=\frac{f^{(k+1)}(\xi_L)}{(k+1)!} x^{k+1}$ and using the bound on the derivatives we get 
$$ |f(x)| \leq \frac{M}{(k+1)!}|x|^{k+1}$$ for all positive integers $k$.
Taking $ k \to \infty$ we get that $f(x)=0$. Since $x$ was arbitrary, the problem is solved.
