Computing $\lim_{n\to\infty} \prod_{k=1}^n(1-\frac{x^2k^{2\alpha}}{n^{2 \alpha+1}})$ Let $\alpha>0,x \in \mathbb{R}$ 
I am having a problem in computing the following limit:
$$\lim_{n \to \infty} \prod_{k=1}^n\bigg(1-\frac{x^2k^{2a}}{n^{2a+1}}\bigg).$$ 
In fact: the problem was to prove that if $(X_n)_n$ is a sequence of i.i.d random variables such that $E[X_1^2]<+\infty,E[X_1]=0$, then $$\frac{1}{n^{\alpha+1/2}}\sum_{k=1}^n k^\alpha X_k$$ converges to a normal distribution, using characteristic functions (without using Lindeberg's condition ), so I imitate the proof of the central limit theorem, and I obtain the above product, hoping that I didn't make any error.
I appreciate any different way to compute it. 
 A: Note that, for any $ x\in\mathbb{R}_{-} $, we have : \begin{aligned}\left|x-\ln{\left(1+x\right)}\right|=x^{2}\int_{0}^{1}{\frac{1-t}{\left(1+xt\right)^{2}}\,\mathrm{d}t}\leq \frac{x^{2}}{\left(1+x\right)^{2}}\int_{0}^{1}{\left(1-t\right)\mathrm{d}t}=\frac{x^{2}}{2\left(1+x\right)^{2}}\end{aligned}
Thus, if $ n $ is a positive integer greater than $ x^{2} $, we have : \begin{aligned}\left|-\frac{x^{2}}{n}\sum_{k=1}^{n}{\left(\frac{k}{n}\right)^{2a}}-\sum_{k=1}^{n}{\ln{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)}}\right|&=\sum_{k=1}^{n}{\left(-\frac{x^{2}k^{2a}}{n^{2a+1}}-\ln{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)}\right)}\\ &\leq\frac{x^{4}}{2n^{4a+2}}\sum_{k=1}^{n}{\frac{k^{4a}}{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)^{2}}}\\ &\leq\frac{x^{4}}{2n\left(1-\frac{x^{2}}{n}\right)^{2}}\times\frac{1}{n}\sum_{k=1}^{n}{\left(\frac{k}{n}\right)^{4a}}\underset{n\to +\infty}{\longrightarrow}0\times\int_{0}^{1}{x^{4a}\,\mathrm{d}x}=0\end{aligned}
(To get the last inequality, we used that $ k\leq n $)
Thus, $$ \lim_{n\to +\infty}{\sum_{k=1}^{n}{\ln{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)}}}=\lim_{n\to +\infty}{\left(-\frac{x^{2}}{n}\sum_{k=1}^{n}{\left(\frac{k}{n}\right)^{2a}}\right)}=-x^{2}\int_{0}^{1}{x^{2a}\,\mathrm{d}x}=-\frac{x^{2}}{1+2a} $$
Since $ \exp $ is a continuous function, $ \prod\limits_{k=1}^{n}{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)}=\exp{\left(\sum\limits_{k=1}^{n}{\ln{\left(1-\frac{x^{2}k^{2a}}{n^{2a+1}}\right)}}\right)}\underset{n\to +\infty}{\longrightarrow}\exp{\left(-\frac{x^{2}}{1+2a}\right)} \cdot $
A: Similar to @CHAMSI's answer
$$P_n= \prod_{k=1}^n\left(1-\frac{x^2k^{2\alpha}}{n^{2 \alpha+1}}\right)\implies \log(P_n)=\sum_{k=1}^n\left(1-\frac{x^2k^{2\alpha}}{n^{2 \alpha+1}}\right)\sim -\sum_{k=1}^n \frac{x^2k^{2\alpha}}{n^{2 \alpha+1}}$$ that is to say
$$\log(P_n) \sim -x^2 n^{-(2 \alpha +1)} H_n^{(-2 \alpha )}$$ Now, using the asymptotics of generalized harmonic numbers
$$\log(P_n) \sim -\frac{x^2}{2 a+1}-\frac{x^2}{2 n}+\cdots$$ and for very large values of $n$
$$P_n \sim \exp\left(-\frac{x^2}{2 a+1}\right)$$
A: Another approach is the following: the use of integral test for convergence (https://en.wikipedia.org/wiki/Integral_test_for_convergence):
Notice that $$\sum_{k=1}^n\ln(1-\frac{x^2(k+1)^{2\alpha}}{2n^{2\alpha+1}}) \leq\int_{1}^{n+1}\ln(1-\frac{x^2y^{2\alpha}}{2n^{2\alpha+1}})dy\leq\sum_{k=1}^n\ln(1-\frac{x^2k^{2 \alpha}}{2n^{2\alpha+1}}),$$
(We can take $\ln$ since for $n$ big enough $1-\frac{x^2y^{2\alpha}}{2n^{2\alpha+1}}>0$,)
We also have $$\int_{1}^{n+1}\ln(1-\frac{x^2y^{2\alpha}}{2n^{2\alpha+1}})dy=\int_{\frac{1}{n+1}}^{1}(n+1) \ln(1-\frac{u^{2\alpha}x^2(n+1)^{2\alpha}}{2n^{2\alpha+1}})du$$ which converges, by the monotone convergence theorem, to $$-\frac{x^2}{2}\int_0^1u^{2\alpha}du=\frac{-x^2}{2(2\alpha+1)}$$ and then the characteristic function will be $e^{-\frac{x^2}{2(2\alpha+1)}}$ ($N(0,\frac{1}{2\alpha+1})$)
(Of course supposing that $E[X^2]=1$ in the above question)
