Logarithmic Limit Theorem I am working on the following:
Let $\{V_n\}$ be a sequence of iid random variables all distributed as $Y=X/ \sqrt{U}$ where $U$ is uniform on $[0,1]$ and $P(X=-1)=P(X=1)=1/2$.  
Prove that $\displaystyle\lim_{n \to \infty} \frac{V_1+\cdots+V_n}{(n\log n)^{1/2}}= N(0,1)$ in distribution.
I know that the distribution of $Y$ is $f_Y(y)=|y|^{-3}1_{|y|>1}$.  I thought maybe this could be worked out using characteristic functions, but I'm not sure how to achieve the standard normal.
 A: Write the characteristic function of $Z_n=\frac{V_1+\cdots+V_n}{(n\log n)^{1/2}}$:
$$
\varphi_{Z_n}(t)=\left(\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)\right)^n.
$$
Next consider the inner function:
$$
\varphi_{V_1}(t)=2\int_1^\infty \dfrac{\cos(ty)}{y^3}\,dy,
$$
$$
\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)=2\int_1^\infty \dfrac{\cos\left(\frac{ty}{\sqrt{n\log n}}\right)}{y^3}\,dy.
$$
Change variables in the above integral $x=\frac{y}{\sqrt{n\log n}}$:
$$
\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)=\frac{1}{n\log n}\int\limits_{\frac{1}{\sqrt{n\log n}}}^\infty \dfrac{2\cos\left(tx\right)}{x^3}\,dx.
$$
Since $\frac{1}{\sqrt{n\log n}}\to 0$, one can split integral to a pair of integrals: over $\left(\frac{1}{\sqrt{n\log n}}, 1\right)$ and over $(1,\infty)$. Note that the second integral is c.f. of $V_1$ and its absolute value is bounded by $1$. So
$$
\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)=\frac{1}{n\log n}\left(\int\limits_{\frac{1}{\sqrt{n\log n}}}^1 \dfrac{2\cos\left(tx\right)}{x^3}\,dx + O(1)\right).
$$
Use Taylor expansion for cosine: $|2\cos(tx)-2+t^2x^2|\leq 2\frac{t^4x^4}{4!}$ since Taylor expansion for cosine is  an alternating series. Then we can substitute $2-t^2x^2$ instead of $2\cos(tx)$ into the above integral. the difference with the initial integral will be $O(1)$.
$$
\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)=\frac{1}{n\log n}\left(\int\limits_{\frac{1}{\sqrt{n\log n}}}^1 \dfrac{2-t^2x^2}{x^3}\,dx + O(1)\right)=\dfrac{n\log n+t^2\log\left(\frac{1}{\sqrt{n\log n}}\right)+O(1)}{n\log n}=
$$
$$
=1-\frac{t^2}{2n}-\dfrac{\frac{t^2}{2}\log\log n+O(1)}{n\log n}=1-\frac{t^2}{2n}+o(1).
$$
Finally, 
$$\varphi_{Z_n}(t)=\left(\varphi_{V_1}\left(\frac{t}{\sqrt{n\log n}}\right)\right)^n=\left(1-\frac{t^2}{2n}+o(1)\right)^n\to e^{-t^2/2}.
$$
