Is sum of infinite exponential functions an entire function?

Question

My question is actually more specific than the title. Let $f_{k}(x)=a_{k}^{x}$ where $\{a_{k};k=1,2\cdots\}$ is a sequence of real numbers in $[0,1)$ satisfying $\sum_{k\geq 1}a_{k}=c<\infty$. For each $k$, $f_{k}(x)$ is entire - Taylor's expansion converges for every $x\in [1,\infty)$. Let $$ F(x)=\sum_{k\geq 1}f_{k}(x).$$ Clearly $F(x)$ is well-defined over domain $x\in [1,\infty)$. What (sufficient) condition is needed for $F(x)$ to be entire on $[1,\infty)$?

Answer 1

As remarked in the comments, the absolute convergence of $\sum_{k\geq 1}a_k$ is not enough to ensure that $F(x)$ is differentiable in a right neighbourhood of $x=1$: if we take $a_k$ as $\frac{1}{(k+1)\log^2(k+1)}$ or as $\frac{1}{(k+1)\log(k+1)\log\log^2(k+1)}$ then $\sum_{k\geq 1}a_k$ is absolutely convergent but $\sum_{k\geq 1}a_k\log(a_k)$ is divergent, always by Cauchy's condensation test. On the other hand, if we assume $a_k=O\left(\frac{1}{k^{1+\varepsilon}}\right)$ for some fixed $\varepsilon > 0$ we have $$ \left|F^{(m)}(1)\right|\leq (1+\varepsilon)^m\sum_{k\geq 1}\frac{\left(\log k\right)^m}{k^{1+\varepsilon}}\ll\frac{1}{\varepsilon}\left(1+\frac{1}{\varepsilon}\right)^m\cdot m! $$ $$\left|F^{(m)}(x)\right|=\left|\sum_{k\geq 1}a_k\left(\log a_k\right)^m e^{(x-1)\log a_k}\right|\ll\left|F^{(m)}(1)\right| $$ hence the radius of convergence of the Taylor series of $F(x)$ at some $x>1$ is $\gg\min(\varepsilon,1)$ and $F(x)$ is analytic over $[1,+\infty)$.