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Consider the following result($d$ denotes the dimensions and $0<t<T$$$c\left(\sum_{j=0}^\infty\frac{\Gamma^j(1-\kappa)}{\Gamma((j+1)(1-\kappa))}t^{j(1-\kappa)-\kappa}\right)^{\frac{1}{2}}\leq c t^{-\frac{\kappa}{2}}, \text{ where } \frac{\kappa}{2}=\frac{d}{\alpha}-\frac{d}{2q\alpha}, \alpha+d\geq \frac{\alpha+d}{2q} ,0<\alpha\leq 2, \text{ and } 2q>1,$$ which appears in the article Quantitative normal approximations for the stochastic fractional heat equation. I notice that it only takes the term of $j=0$. I try to use the conclusion $\Gamma(z)\sim z^{z-1/2}e^{-z}$, and prove it by considering the rapid decrease. But I don't know if it is true and how to explain it convincingly.

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    $\begingroup$ To simplify the question, why don't you divide by $c$, square, and multiply by $t^\kappa$ both sides? And maybe call $s:=t^{\kappa-1}$. $\endgroup$ Commented Jul 10 at 17:41
  • $\begingroup$ Also, $\alpha, d, q, t$ are completely useless as the information is not enough. Why don't you just cut them from the question and say in what range you want $\sigma:=1-\kappa $ and $\tau:=\Gamma(1-\kappa)t^{1-\kappa}$? $\endgroup$ Commented Jul 10 at 17:52
  • $\begingroup$ So the question reads: is $\sum_{j=0}^\infty\frac{\tau^j}{\Gamma((j+1)\sigma)}\leq1$ for $\tau$,$\sigma$ in the prescribed range? $\endgroup$ Commented Jul 10 at 18:00
  • $\begingroup$ Thank you so much Professor Pietro Majer for your advice. But the corresponding series is not geometric progression as I expected. How can I get its bound ? $\endgroup$
    – Y. Li
    Commented Jul 12 at 7:22
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    $\begingroup$ Note that for $\sigma=1$ you get $e^\tau$. In general you may use the Stirling formula $\endgroup$ Commented Jul 12 at 8:42

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