*Note* The following strictly speaking does not answer the question but it may answer what the OP *meant*, i.e., under which conditions a formal power series defines a function. **Edit** My memory didn't fail me. The following is taken almost verbatim from Reed and Simon Vol 4 see [here](https://books.google.it/books?id=dnjNCgAAQBAJ&pg=PA43&lpg=PA43&dq=strong+asymptotic+series&source=bl&ots=nyxqxv4tsQ&sig=bKz17Snu06FKc7gqwOOUIWzARrc&hl=en&sa=X&ved=0ahUKEwjnqobmtLTYAhUINxQKHTMBCuEQ6AEIODAC#v=onepage&q=strong%20asymptotic%20series&f=false) > **Definition 1** we say that a function $E\left(\lambda\right),$ analytic in a sectorial region $$\Omega=\left\{ z|0<\left|z\right|<B;\left|\textrm{arg}\left(z\right)\right|<\pi/2+\epsilon\right\} , $$ obeys a *strong asymptotic condition* and has $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as *strong asymptotic series* (SAS) if there are positive constant $C$ and $\sigma$ such that $$ \left|E\left(\lambda\right)-\sum_{n=0}^{N}a_{n}\lambda^{n}\right|<C\sigma^{N+1}\left(N+1\right)!\left|\lambda\right|^{N+1} $$ for all $N$ and all $\lambda\in\Omega$. Given the above one has: > **Theorem** A SAS defines a function in the sense that if two analitic functions $f,g$ have the same SAS then $f=g$. *Remark:* Informally the coefficients must not grow too fast. In fact SAS implies $\left|a_{n}\right|\le C \sigma^{n}n!$. There are series associated with simple examples for which $a_n$ behaves like $(kn)!$ with $k > 1$. Thus, a strong asymptotic condition cannot hold in such cases. However with a simple modification even this case may be treated. This suggests that we define: > **Definition 2** we say that a function $E\left(\lambda\right),$ analytic in a sectorial region $$\Omega=\left\{ z|0<\left|z\right|<B;\left|\textrm{arg}\left(z\right)\right|<\pi/2+\epsilon\right\} ,$$ obeys a **modified strong asymptotic condition of order $k$** and has $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as an **order $k$ strong asymptotic series** if there are positive constant $C$ and $\sigma$ such that $$ \left|E\left(\lambda\right)-\sum_{n=0}^{N}a_{n}\lambda^{n}\right|<C\sigma^{N+1}\left [ k(N+1) \right]!\left|\lambda\right|^{N+1} $$ for all $N$ and all $\lambda\in\Omega$. The above result extends to this case too. Namely > **Theorem 2** If $f,g$ both have $\sum_{n=0}^{\infty}a_{n}\lambda^{n}$ as order $k$ strong asymptotic series, then $f=g$.