Let $\{ \delta_n (\varepsilon) \}$ be an asymptotic sequence as $\varepsilon \rightarrow 0$, and $a_n$ be independent of $\varepsilon$ and $\delta_n (\varepsilon)$. A series $\sum \limits_{n=0}^{\infty} a_n \delta_n (\varepsilon)$ is said to be an asymptotic expansion of a function $f(\varepsilon)$ as $\varepsilon \rightarrow 0$ if for all $N>1$, $f(\varepsilon)-\sum \limits_{n=0}^{N} a_n \delta_n (\varepsilon)=o(\delta_N (\varepsilon))$ as $\varepsilon \rightarrow 0$.
Write down an asymptotic expansion for the function $$f(\varepsilon)=\cos(2\varepsilon)$$ in the limit $\varepsilon \rightarrow 0^+$.
Following the definition, I need $$\frac{\cos(2\varepsilon)-\sum \limits_{n=0}^N a_n \delta_n (\varepsilon)}{\delta_N (\varepsilon)} \rightarrow 0$$as $\varepsilon \rightarrow 0^+$
so I need to choose my $a_n$ and $\delta_n(\varepsilon)$ so that this is satisfied, right?
But how can I do this because the numerator is always going to have the negative of the largest $N$ term as the denominator.
Please help.
For finite limit points, convergent power series in the neighbourhood of the point are examples of asymptotic series.
For a bit more detail,
If $$f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n$$ converges then for each $N$, $$f(x) = \sum_{n=0}^N a_n (x-x_0)^n + R_N$$ where $$R_N = (x-x_0)^{N+1}\left(\sum_{n=N+1}^\infty a_n (x-x_0)^{n-(N+1)}\right)$$ is bounded so $$R_N \in o((x-x_0)^N) \text{ as } x \to x_0$$