Asymptotics de $\prod_{x=1}^{\lceil\frac{n}{\log_2{n} }\rceil} \left(\frac{1}{\sqrt{n}} + x\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right)\right) $

Question

Estoy tratando de trabajar fuera de la gran $n$ asymptotics de $$S_n = \prod_{x=1}^{\lceil\frac{n}{\log_2{n} }\rceil} \left(\frac{1}{\sqrt{n}} + x\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right)\right) .$$

Here is my attempt so far $$\prod_{x=1}^{k} (A + Bx) = \frac{B^k \Gamma(k+1+A/B)}{\Gamma(1+A/B)}.$$

In our case $A/B \approx \sqrt{n}$ and $\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right) \approx \frac{1}{n}$. Therefore

$$S_n \approx \frac{\frac{1}{n}^{\frac{n}{\log_2{n}}} \left(\frac{n}{\log_2{n}} + \sqrt{n}\right)!}{\sqrt{n}!}$$

I am not really sure where to go from here, if I haven't already taken an approximation too far.

---

I tried taking logs and defining in maple

f:=(x,n)-> log(1/sqrt(n)+x*(1/n-2/n^(3/2)))

Now if you do

plot(-(sum(f(x, n), x = 1 .. n/log(n))), n = 10 .. 100)

you get what looks like a linear function of $n$. However if you do

limit(-(sum(f(x, n), x = 1 .. n/log(n)))/n, n = infinity)

you get $0$.

Answer 1

Como se sugiere en los comentarios, utilice el Stirling fórmula: \begin{align} \ln S_n&=k\ln B+\ln \Gamma(1+k+A/B)-\ln \Gamma(1+A/B)\approx\\ &\approx k\ln B+ \left(k+\frac AB\right)\left[\ln\left(k+\frac AB\right)-1\right]+\frac12\ln\left(k+\frac AB\right)\\&\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB\approx \\ &\approx \color{red}{k\ln B}+\left(\color{red}{k}+\frac AB\right)\left[\color{red}{\ln k} +\frac{A}{Bk}-\frac12\left(\frac{A}{Bk}\right)^2-\color{red}{1}\right]+\frac12\ln k\\ &\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB. \end{align} Ahora tenemos \begin{align} &B=\frac1n-\frac{2}{n^{3/2}} \qquad \Longrightarrow \qquad \ln B\approx -\ln n-\frac{2}{\sqrt n}-\frac2n+O\left(n^{-3/2}\right),\\ & \frac AB=\frac{1}{\sqrt n}/\left(\frac1n-\frac{2}{n^{3/2}} \right)\approx \sqrt n+2+O\left(n^{-1/2}\right),\\ &\ln\frac AB\approx \frac12\ln n+\frac{2}{\sqrt n}+O\left(n^{-1}\right). \end{align} Esto en principio es suficiente para determinar la asymptotics de $S_n$$o(1)$ -, pero sólo voy a escribir un par de líderes (que se desprenden de los que se muestran arriba en rojo): $$\ln S_n\approx -n\frac{\ln\log_2n +1}{\log_2n}+O\left(n^{\frac12}\ln n\right).$$