Let $f\in C^{\infty}(\mathbb{R})$ verifying,
$a)$ exists $L>0$ : $\forall x\in \mathbb{R}$ and $\forall n\ge 1$
$$|f^{(n)}(x)| \le L,$$
$b)$ $$f\left(\frac{1}{n}\right)=0 \quad \forall n\ge 1.$$
Prove that, $$f(x)\equiv 0, \quad \mbox {on} \quad \mathbb{R}.$$
On
Method 1. The condition $$ \lvert\,f^{(n)}(x)\rvert\le L, \quad\text{for all}\,\,x\in\mathbb R\,\,\,\text{and}\,\,\,n\in\mathbb N, \tag{1} $$ implies that the Taylor series of $f$ converges for all $x\in\mathbb R$, and hence $f$ is equal to $$ f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)\,x^n}{n!}, $$ for all $x\in\mathbb R$.
As $\,f\left(\dfrac{1}{n}\right)=0$, then $f(0)=0$, because of continuity.
Assume now that $\,f\not\equiv 0$. In case, there would be a $k\in\mathbb N$, such that $\,f^{(k)}(0)\ne 0$ while $\,f^{(0)}(0)=\cdots =f^{(k-1)}(0) $. Then $$ 0=n^kf\left(\frac{1}{n}\right)=n^k\sum_{j=k}^\infty \frac{f^{(j)}(0)}{j!\,n^{\,j}}=\frac{f^{(k)}(0}{k!}+\sum_{n=k+1}^\infty \frac{f^{(j)}(0)}{j!\,n^{\,j-k}}\to \frac{f^{(k)}(0)}{k!}, $$ as $n\to\infty$, and hence $\dfrac{f^{(k)}(0)}{k!}=0$, which contradicts our assumption. Thus $f\equiv 0$.
Method 2. The condition $(1)$ implies that $\,f$ is real-analytic in $\mathbb R$, and hence it can be expressed locally (in fact, globally) as a power series.
The fact that $f$ is identically zero is a consequence of the following fact:
If $f,g:(a,b)\to\mathbb R$ are real-analytic and coincide in a set with has an accumumation point in $(a,b)$, then $f\equiv g$.
In our case, $f$ coincides with $g\equiv 0$ at the points $\left\{\frac{1}{n}:n\in\mathbb N\right\}$, which have $0$ as an accumulation point.
Hints. (a) Note that the Taylor series for $f$ at 0, that is $$ \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n $$ converges to $f$ everywhere (use Taylor's theorem and the boundedness of the derivatives).
(b) Hence, both powerseries $\sum_n \frac{f^{(n)}(0)}{n!}x^n$ and $0$ agree on an infinite set with an accumulation point. That is, they are equal.