# Is there a condition that a power series is $O(1)$?

There exists a power series $$f(x) = \sum\limits_{n=1}^\infty a_n x^n = a_0 +a_1 x + a_2 x^2 + \cdots$$ Assume that the radius of convergence of this power series is $\infty$, i.e.,

$$\limsup_{n\rightarrow\infty} \big(|a_n|^{\frac{1}{n}}\big) = 0$$ by Cauchy-Hadamard theorem.

My question :

"Is there a sufficient (or necessary and sufficient) condition that $f(x)=O(1)$ as $x \rightarrow \infty$?"

For example,

the Maclaurin series of $\sin x$, $\cos x$, and $e^{-x}$ are well-defined in $\mathbb{R}$, and $O(1)$ as $x \rightarrow \infty$.

• I don't know the answer for $\mathbb{R}$, but if you ask the same question on $\mathbb{C}$, the necessary and sufficient condition for the boundedness of a power series with infinite convergence radius is that it is constant – Max Jan 5 '18 at 19:52

A necessary and sufficient condition comes from Ramanujan's master theorem.

Let $z = x+ iy$. If and only if there is a holomorphic function $\phi(z)$ for $\Re(z) > 0$ and $|\phi(z)| < C e^{\rho|x| + \kappa|y|}$ for $C,\rho \in \mathbb{R}^+$ and $0<\kappa < \pi/2$ then

$$f(x) = \sum_{n=0}^\infty (-1)^n \phi(n+1)\frac{x^n}{n!}$$ is $O(1)$ for $|\arg(x)| < \kappa$ as $x \to \infty$.

It's a little tricky to prove this, it's rather complicated. Don't have time to find a reference at the moment.

• Could you explain how $f(x)$ is $O(1)$ with you assumption? – Doyun Nam Jan 6 '18 at 15:38
• It comes from the isomorphism of spaces (holomorphic functions bounded as I wrote, and the space of $O(1)$ functions) induced by the Mellin transform $\Gamma(1-z)\phi(z) = \mathcal{M}f = \int_0^\infty f(x)x^{-z} \, dx \Leftrightarrow f(x) = \sum_{k=0}^\infty (-1)^k \phi(k+1)\frac{x^k}{k!}$. What I wrote is slightly incorrect though (wrote it a bit too fast), this is a sufficient and necessary for $f(x) = O(1/x)$. We can make $f(x)$ grow like any power of $x$ by noticing if $\phi(z)$ holomorphic for $\Re(z) > a$ (bounded as I wrote, $a<1$) then $f(x) = O(1/x^{1-a})$ – user335907 Jan 7 '18 at 0:54
• So, for example, if $f$ is $O(1)$ then there exists a function $\phi$ such that $\phi(z)\Gamma(1-z) = \mathcal{M}f$ where $(-1)^n\phi(n+1)/n! = a_n$. (We take $\mathcal{M}$ to be an extension of the Mellin transform, not simply the integral which could diverge). And if we have $\phi$ we can get $f$. – user335907 Jan 7 '18 at 0:55