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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 $x$.

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]. I know this relationships holds true when $a(k):=k$ (see [2], [3]), 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.

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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 $x\to\infty$.

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  • $\begingroup$ Thank you for your answer. I've amended the question to make sure that $a(t)$ is smooth. Is this allowed, or should I create a new question? $\endgroup$
    – Dispersion
    Commented Jan 4, 2023 at 21:44
  • $\begingroup$ (i) The smoothness will not help, because the step-function $a$ can be approximated arbitrarily closely in $L^1$ by a smooth function, without changing the values of any of the $a(k)$'s and only negligibly changing the values of any of the integrals $\int_k^{k+1}\frac{x^t}{\Gamma(1+a(t))}\,dt$, for any $k=0,1,\dots$. (ii) Indeed, changing the question in a way that would invalidate a valid answer should be avoided. $\endgroup$
    – Iosif Pinelis
    Commented Jan 4, 2023 at 22:31
  • $\begingroup$ I see. Do you know conditions that could be placed on $a(t)$ such that the asymptotic relation does indeed hold between integral and series? $\endgroup$
    – Dispersion
    Commented Jan 4, 2023 at 22:48
  • $\begingroup$ @Zachary : I guess it would be enough if $a$ can be extended to an analytic function on the set $\{z\in\mathbb C\colon\Re z>0\}$ -- but that would certainly be another question. $\endgroup$
    – Iosif Pinelis
    Commented Jan 4, 2023 at 22:57
  • $\begingroup$ Wouldn’t it make more sense for it to still be under this question? I still don’t think the question has been fully answered with all due respect, since the question asks what conditions are necessary for the given asymptotic relation to hold, not merely the existence of a counter-example. I wouldn’t mind posting another question but it seems to me that it would just be a duplicate. $\endgroup$
    – Dispersion
    Commented Jan 5, 2023 at 1:34

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