# Comparison power series

I have a doubt concerning concerning power series comparison.

There are two power series $$f$$ and $$g$$ with the coefficients $$(a_i)_{i \in \mathbb{N}}$$ and $$(b_i)_{i \in \mathbb{N}}$$. That is saying :

$$f(t) = \sum_{i=0}^\infty a_i t^i$$

and same thing for $$g$$ with the $$b_i$$. We know that both $$f$$ and $$g$$ are greater than $$0$$ on $$\mathbb{R}_+$$.

Can we say that if $$\forall i, |a_i| > |b_i|$$,so we have $$\forall t > 0, f(t) > g(t)$$ ?

Could anyone answer this question by a reference or a proof ?

If it is not true, what conditions let this sentence be checked ?

No that's not enough. Consider the power series $$\sum_{i = 0}^\infty (-1)^i\frac{2^i}{i!} t^i$$ and $$\sum_{i = 0}^\infty \frac{1}{i!} t^i$$ of $$e^{-2t}$$ and $$e^t$$. Obviously $$|(-1)^i\frac{2^i}{i!}| = \frac{2^i}{i!} > \frac{1}{i!} = |\frac{1}{i!}|$$ but $$e^{-2t} < e^t$$ for all $$t > 0$$.
• @Gaetano Just omit the absolute values around the coefficients: $a_i > b_i$. Apr 23, 2023 at 22:10
• Ok, I can say that we have $|a_k|>|b_k|$ for all $k$ and $sign(a_k) = sign(b_k)$ for all $k$ too. Can We conclude from that ? @ayeayemaung Apr 25, 2023 at 13:03