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 ?

Thank you in advance !


1 Answer 1


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 $t > 0$.

  • $\begingroup$ Thank you very much ! So how can we strengthen the hypothesis on the coefficient so that the inquality is verified ? $\endgroup$
    – NancyBoy
    Apr 23, 2023 at 15:34
  • $\begingroup$ @Gaetano Just omit the absolute values around the coefficients: $a_i > b_i$. $\endgroup$
    – balddraz
    Apr 23, 2023 at 22:10
  • $\begingroup$ Thank you ayeayemaung, however, I want to compare to series expansion with information on the absolute value of the coefficients... $\endgroup$
    – NancyBoy
    Apr 24, 2023 at 15:22
  • $\begingroup$ @Gaetano that's probably not possible based on absolute values alone; you need some information about the sign of the coefficients too $\endgroup$
    – balddraz
    Apr 24, 2023 at 16:29
  • $\begingroup$ 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 $\endgroup$
    – NancyBoy
    Apr 25, 2023 at 13:03

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