Let $N \geqslant 0$ an integer and $(u_n)_{n \geqslant 0}$ be the sequence defined as follow: $$u_n := \int _0^{+ \infty} e^{-nt} \left( \sum_{k=0}^{N-1} \frac{t^k}{k!}\right)^n\,\mathrm{d} t.$$
Find an explicit constant $C_N$ only depending on $N$ (explicit but not necessarily computable, e.g $\displaystyle \int _{\mathbb{R}} e^{-u^2} \, \mathrm{d}u$ is such a constant) such that : $$u_n \leqslant \frac{C_N}{n^{1/N}}$$ for all
For $N=1$ one gets $C_1=1$. The real problem starts with $N=2$: I use the change of variable $u=\sqrt{n}t$ which allow us to write :
$$v_n:= \sqrt{n}\int_1 ^{+\infty} e^{-nt} \left( 1+t\right)^n\,\mathrm{d} t = \int_0 ^{1} e^{-\sqrt{n}u +n \log \left(1+\frac{u}{\sqrt{n}}\right)}\,\mathrm{d} t = \int_0 ^{1} e^{-u^2/2 + \sum_{k=3}^{+\infty} (-1)^k \frac{u^k}{kn^{k/2-1}}}\,\mathrm{d} t .$$
Then, Lebesgue dominated convergence theorem enable us to write that $v_n \to \int_0^{+ \infty} e^{-u^2} \, \mathrm{d}u$. For the remaining part of the integral defining $u_n$ one can perform other methods (using the concavity of the logarithm). This shows that the asymptotic comportement is the one expected but does not gives an explicit global bound for $\sqrt{n} u_n$ !
If anyone knows how to find such bounds, please feel free to answer, even if you think that your method is not optimal.
Thanks,