Solution. The coefficients $c_{n}$ necessarily coincide with

$$\frac{\varphi^{(n)}(0)}{n!}=\frac{i^{n} E X^{n}}{n!}$$

where $X$ is a random variable with characteristic function $\varphi$. Therefore, the series

$$\sum_{k \geqslant 0} \frac{i^{k} \mathrm{E} X^{k}}{k!} t^{k}$$

converges on $\mathbb{R}$. Moreover, for all $n \geqslant 1$ we have

$$\varphi(t)=\sum_{k=0}^{2 n-1} \frac{i^{k} \mathrm{E} X^{k}}{k!} t^{k}+R_{n}(t) t^{2 n}$$

where

$$|R(t)| \leqslant \frac{3 \mathrm{E} X^{2 n}}{(2 n)!}, \quad t \in \mathbb{R}$$

Thus, by taking the limit as $n \rightarrow \infty$, we obtain that on $\mathbb{R}$ the relation

$$\varphi(t)=\sum_{k \geqslant 0} \frac{i^{k} E X^{k}}{k!} t^{k}=\sum_{n \geqslant 0} c_{n} t^{n}=f(t)$$

holds.