Let $P(X_j=j)=P(X_j=-j)=1/2j^{\beta}$ and $P(X_j=0)=1-j^{-\beta}$ where $\beta\in(0,1)$, then $S_n/n^{(3-\beta)/2)}\Rightarrow c\chi$ 
Suppose $P(X_j=j)=P(X_j=-j)=1/2j^{\beta}$ and $P(X_j=0)=1-j^{-\beta}$, where $\beta>0$. Show that:
(i) If $\beta>1$ then $S_n\to S_\infty$ a.s.
(ii) If $\beta\in(0,1)$ then $S_n/n^{(3-\beta)/2}\Rightarrow c\chi$.
(iii) If $\beta=1$ then $S_n/n\Rightarrow\aleph$, where
$$E\exp(it\aleph)=\exp\left(-\int_0^1 x^{-1}(1-\cos(xt)\,\mathrm{d}x\right).$$

This is problem 3.4.13 in Durrett's Probability text, part (i) was rather trivial, I feel fine about that part. I am having a difficult time on part (ii) though and would like verification for part (iii).
My ideas so far for part (ii) is to define the triangular array as $S_{n,m}=\dfrac{X_m}{n^{(3-\beta)/2}}$, and then use the Lindeberg-Feller theorem, but I am getting hung up on the details.
For part (iii) consider:
It is a well-known theorem of Levy that if $\{X_n\}$ is a collection of random variables and $Y$ is another random variable then $X_n \Rightarrow Y$ iff $\phi_{X_n}(t) \rightarrow \phi_Y(t)$ as $n \rightarrow \infty$ and $\phi_Y$ is continuous at $t = 0$.  Moreover, by properties of Fourier transforms, $\phi_{S_n/n}(t) = \prod\limits_{1 \leq j \leq n} \phi_{X_j/n}(t)$. Now,
$$\phi_{X_j/n}(t) = \int_{\mathbb{R}} \mathrm{d}\lambda e^{it\lambda} \mathbb{P}\left(\frac{X_j}{n} = \lambda\right) = 1-\frac{1}{j} + \frac{1}{2j}(e^{it\frac{j}{n}} + e^{-it\frac{j}{n}}) = 1-\frac{1}{j}(1-\cos(tj/n)).
$$
This is clearly real-valued and positive, so that we can write
$$
\log\phi_{S_n/n}(t) = \sum_{j = 1}^n \log\left(1-\frac{1}{n}\cdot \frac{n}{j}(1-\cos(tj/n)\right),
$$
so, up to an $O(1/n)$ error term, we have
$$
\log \phi_{S_n/n}(t) = \frac{1}{n}\sum_{j=1}^n \frac{n}{j}(1-\cos(tj/n)) + O\left(\frac{1}{n}\right).
$$
The sum on the right side is a Riemann sum for the exponential, so taking $n \rightarrow \infty$, we get $\phi_{S_n/n}(t) \rightarrow E\left(e^{it\aleph}\right)$, in our notation, the latter of which is continuous at $0$.
 A: $\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\d{\mathrm{d}}$As is written at the start of exercise section, $X_1, X_2, \cdots$ are independent.
Define $X_{n, k} = \dfrac{X_k}{n^{\frac{3 - β}{2}}}$ for $1 \leqslant k \leqslant n$. Since Lindeberg's condition does not apply for $\{X_{n, k} \mid 1 \leqslant k \leqslant n\}$, so the proposition has to be proved directly. Since$$
φ_{n, k}(t) := E(\exp(\i t X_{n, k})) = \frac{1}{k^β} \cos\frac{kt}{n^{\frac{3 - β}{2}}} + \left( 1 - \frac{1}{k^β} \right), \quad \forall t \in \mathbb{R}
$$
it suffices to prove that there exists a constant $c$ that$$\lim_{t → ∞} \prod_{k = 1}^n φ_{n, k}(t) = \exp\left( -\frac{1}{2} c^2 t^2 \right). \quad \forall t \in \mathbb{R}
$$
For a fixed $t$, in order to apply Exercise 3.1.1., denote $c_{n, k} = φ_{n, k}(t) - 1 = \dfrac{1}{k^β} \left( \cos\dfrac{kt}{n^{\frac{3 - β}{2}}} - 1 \right)$, it suffices to prove that$$
\lim_{n → ∞} \max_{1 \leqslant k \leqslant n} |c_{n, k}| = 0, \quad \lim_{n → ∞} \sum_{k = 1}^n c_{n, k} = -\frac{1}{2} c^2 t^2, \quad \sup_{n \geqslant 1} \sum_{k = 1}^n |c_{n, k}| < +∞.
$$
Since $|c_{n, k}| \leqslant \dfrac{1}{k^β} · \dfrac{1}{2} \left( \dfrac{kt}{n^{\frac{3 - β}{2}}} \right)^2 = \dfrac{k^{2 - β} t^2}{2n^{3 - β}} \leqslant \dfrac{t^2}{2n}$, then $\lim\limits_{n → ∞} \max\limits_{1 \leqslant k \leqslant n} |c_{n, k}| = 0$ and$$
\sum_{k = 1}^n |c_{n, k}| \leqslant \sum_{k = 1}^n \frac{k^{2 - β} t^2}{2n^{3 - β}} \leqslant \frac{t^2}{2n^{3 - β}} \int_1^{n + 1} x^{2 - β} \,\d x \leqslant \frac{t^2}{2(3 - β)} \left( \frac{n + 1}{n} \right)^β,
$$
which implies $\sup\limits_{n \geqslant 1} \sum\limits_{k = 1}^n |c_{n, k}| < +∞$.
Now, since $\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + o(x^5)\ (x → 0)$, there exists $δ > 0$ such that$$
1 - \frac{x^2}{2} < \cos x < 1 - \frac{x^2}{2} + \frac{x^4}{23}. \quad \forall |x| < δ
$$
For $n > \left( \dfrac{t}{δ} \right)^{\frac{2}{1 - β}}$,\begin{align*}
\sum_{k = 1}^n c_{n, k} &\leqslant \sum_{k = 1}^n \frac{1}{k^β} \left( -\frac{k^2 t^2}{2n^{3 - β}} + \frac{k^4 t^4}{23n^{2(3 - β)}} \right) = -\sum_{k = 1}^n \frac{k^{2 - β} t^2}{2n^{3 - β}} + \sum_{k = 1}^n \frac{k^{4 - β} t^4}{23n^{2(3 - β)}}\\
&\leqslant -\frac{t^2}{2n^{3 - β}} \int_0^n x^{2 - β} \,\d x + n · \frac{n^{4 - β} t^4}{23n^{2(3 - β)}} = -\frac{t^2}{2(3 - β)} + \frac{t^4}{23n^{1 - β}},
\end{align*}$$
\sum_{k = 1}^n c_{n, k} \geqslant -\sum_{k = 1}^n \frac{1}{k^β} · \frac{k^2 t^2}{2n^{3 - β}} \geqslant -\frac{t^2}{2(3 - β)} \left( \frac{n + 1}{n} \right)^β,
$$
thus $\lim\limits_{n → ∞} \sum\limits_{k = 1}^n c_{n, k} = -\dfrac{t^2}{2(3 - β)}$. Applying Exercise 3.1.1., $\dfrac{S_n}{n^{\frac{3 - β}{2}}} \Rightarrow cχ$, where $c = \dfrac{1}{\sqrt{3 - β}}$.
