I have a function defined by its Taylor series $f(x) = \sum_{n=0}^\infty a_n x^n$. The convergence radius is infinite. I am interested in the behavior of $f(x)$ as $x \to \infty$.
I thought about comparing $f$ to a different known function $g$. If $\lim_{x\to \infty} \frac{f(x)}{g(x)} = 1$ then $f(x)$ and $g(x)$ are asymptotically equivalent. Now I can also expand $g$ into a series $g(x) = \sum_{n=0}^\infty b_n x^n$.
This gives $\lim_{x\to \infty} \frac{f(x)}{g(x)} = \lim_{x\to \infty} \frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n}$. Intuitively since $x$ goes to $\infty$ only the terms for $n\to \infty$ should contribute to the limit. In other words if for huge $n \to \infty$ both coefficient are asymptotically equal $a_n \sim b_n$, then also $f$ and $g$ should be asymptotically equal.
This can also be motivated by looking at finite polynomials. Here the limit $\lim_{x\to \infty} \frac{\sum_{n=0}^N a_n x^n}{\sum_{n=0}^N b_n x^n}$ is simply given by $\frac{a_N}{b_N}$. Therefore I would expect that for infinite series $\lim_{x\to \infty} \frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} = \lim_{n\to \infty} \frac{a_n}{b_n}$, which is the same as above.
Is the reasoning and the last formula correct? If not, is there any way to deduce the behavior of a Taylor series for $x \to \infty$?
I think I found an answer myself. However it works only for positive Taylor coefficients (This means that the series grows faster than any polynomial).
Assume that $a_n \sim b_n$ for $n \to \infty$ (or $\lim_{n\to\infty} \frac{a_n}{b_n} = 1$). It implies that for all $\varepsilon > 0$ exists $n_0$ s.t. for $n \geq n_0$: $|a_n - b_n| < \varepsilon b_n$.
If $n_0 = 0$ we can do a simple estimate: $$\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} \leq \frac{\sum_{n=0}^\infty (|a_n-b_n|+b_n) x^n}{\sum_{n=0}^\infty b_n x^n} < \frac{\sum_{n=0}^\infty (\varepsilon b_n +b_n) x^n}{\sum_{n=0}^\infty b_n x^n} = 1 + \varepsilon$$
If $n_0 > 0$ we have $$\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} = \frac{a_0 + \ldots + a_{n_0-1}x^{n_0-1} + x^{n_0} \sum_{n=0}^\infty a_{n+n_0} x^n}{b_0 + \ldots + b_{n_0-1}x^{n_0-1} + x^{n_0} \sum_{n=0}^\infty b_{n+n_0} x^n} = \frac{a_0\frac{1}{x^{n_0}} + \ldots + a_{n_0-1}\frac{1}{x} + \sum_{n=0}^\infty a_{n+n_0} x^n}{b_0\frac{1}{x^{n_0}} + \ldots + b_{n_0-1}\frac{1}{x} + \sum_{n=0}^\infty b_{n+n_0} x^n}$$ Taking the limit for $x \to \infty$ gives $$\lim_{x\to\infty}\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} = \lim_{x\to\infty} \frac{\sum_{n=0}^\infty a_{n+n_0} x^n}{\sum_{n=0}^\infty b_{n+n_0} x^n} < 1 + \varepsilon.$$ Since this is true for all $\varepsilon$: $$\lim_{x\to\infty}\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} \leq 1$$ Also we can exchange both series which gives $$\lim_{x\to\infty}\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} = 1$$ or $\sum_{n=0}^\infty a_n x^n \sim \sum_{n=0}^\infty b_n x^n$ for $x \to \infty$
Therefore in case that all $a_n,b_n \geq 0$ and $\lim_{n\to\infty} \frac{a_n}{b_n}$ exists then $$\lim_{x\to\infty}\frac{\sum_{n=0}^\infty a_n x^n}{\sum_{n=0}^\infty b_n x^n} = \lim_{n\to\infty} \frac{a_n}{b_n}$$