First, some notation. I'll write $f(x)=o(g(x))$ if $\lim\_{x\to\infty} \left|\frac{f(x)}{g(x)}\right|=0$. I'll also write $g(x)=\omega(f(x))$ if $f(x)=o(g(x))$, *i.e.* $\limsup\_{x\to\infty} \left|\frac{g(x)}{f(x)}\right|=\infty$. I'll say $f(x)\sim g(x)$ as $x\to\infty$ if $f(x)=g(x)+o(g(x))$ for all large enough $x$, *i.e.* $f(x)/g(x)=1+o(1)$ for all large enough $x$.




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I'm interested in whether, given a certain class of sequence $a(k)$ (see below), we have that
$$\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}\sim\int\_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt$$
as $x\to\infty$. The sequence $a(k)$ is non-negative, increasing, and approaches $+\infty$ as $k\to\infty$ fast enough that the series on the left hand side converges for every real number $x\in[0,\infty)$—it turns out that this is equivalent to $a(k)=\omega\left(\frac{k}{\log k}\right)$ by [[1]](https://math.stackexchange.com/questions/4611180/characterizing-sequences-ak-satisfying-frac1k-ak-log-ak-to-infty/4611182#4611182). I know this relationships holds true when $a(k):=k$ (see [[2]](https://math.stackexchange.com/questions/4608789/asymptotics-of-int-0-infty-fracx2z-gamma1z-dz?noredirect=1#comment9718521_4608789), [[3]](https://mathoverflow.net/questions/437561/asymptotics-of-int-0-infty-fracx2z-gamma1z-dz-for-large-x/437601#437601)), in which case the series is simply $e^x$, as well as when $a(k):=2k$, in which case the series is $\cosh(\sqrt{x})$.


It seems tempting to use Euler-Maclaurin summation, but the resulting difference between the sum and the integral already ends up being a divergent series for large enough $x$ in the case $a(k):=k$.


Edit: Given the answer below, I require that $a(t)\in C^\infty$ on its domain $[0,\infty)$ and the sequence $a(k)$ is defined to be the restriction $a(k):\mathbb{N}\_0\to[0,\infty)=a(t)|\_{t\in\mathbb{N}\_0}$.


This question is cross-posted at [MSE](https://math.stackexchange.com/questions/4611731/when-is-it-true-that-sum-k-ge-0-fracxk-gamma1ak-sim-int-0-infty?noredirect=1#comment9719961_4611731).



When is it true that $\sum_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}\sim\int_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt$ as $x\to\infty$?
This is not true without further regularity assumptions on $a$.


Indeed, take any sequence $(a(k))\_{k\ge0}$ as in your post and then extend it to the function $a$ on $[0,\infty)$ by the formula $a(t):=a(\lfloor t\rfloor)$. Then for $x>1$ (say)
$$\int\_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt
=\frac{x-1}{\ln x}\,\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}
\not\sim\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}$$
as $x\to\infty$.


