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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 $S_nto S_infty$ a.s.

(ii) If $betain(0,1)$ then $S_n/n^{(3-beta)/2}Rightarrow cchi$.

(iii) If $beta=1$ then $S_n/nRightarrowaleph$, where
$$Eexp(italeph)=expleft(-int_0^1 x^{-1}(1-cos(xt),mathrm{d}xright).$$

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, $phi_{S_n/n}(t) = prodlimits_{1 leq j leq n} phi_{X_j/n}(t)$. Now,
$$phi_{X_j/n}(t) = int_{mathbb{R}} mathrm{d}lambda e^{itlambda} mathbb{P}left(frac{X_j}{n} = lambdaright) = 1-frac{1}{j} + frac{1}{2j}(e^{itfrac{j}{n}} + e^{-itfrac{j}{n}}) = 1-frac{1}{j}(1-cos(tj/n)).
$$
This is clearly real-valued and positive, so that we can write
$$
logphi_{S_n/n}(t) = sum_{j = 1}^n logleft(1-frac{1}{n}cdot frac{n}{j}(1-cos(tj/n)right),
$$
so, up to an $O(1/n)$ error term, we have
$$
log phi_{S_n/n}(t) = frac{1}{n}sum_{j=1}^n frac{n}{j}(1-cos(tj/n)) + Oleft(frac{1}{n}right).
$$
The sum on the right side is a Riemann sum for the exponential, so taking $n rightarrow infty$, we get $phi_{S_n/n}(t) rightarrow Eleft(e^{italeph}right)$, in our notation, the latter of which is continuous at $0$.

$defe{mathrm{e}}defi{mathrm{i}}defd{mathrm{d}}$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$$
φ_{n, k}(t) := E(exp(i t X_{n, k})) = frac{1}{k^β} cosfrac{kt}{n^{frac{3 - β}{2}}} + left( 1 - frac{1}{k^β} right), quad forall t in mathbb{R}
$$
it suffices to prove that there exists a constant $c$ that$$lim_{t → ∞} prod_{k = 1}^n φ_{n, k}(t) = expleft( -frac{1}{2} c^2 t^2 right). quad forall t in mathbb{R}
$$

For a fixed $t$, in order to apply Exercise 3.1.1., denote $c_{n, k} = φ_{n, k}(t) - 1 = dfrac{1}{k^β} left( cosdfrac{kt}{n^{frac{3 - β}{2}}} - 1 right)$, 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 $limlimits_{n → ∞} maxlimits_{1 leqslant k leqslant n} |c_{n, k}| = 0$ 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 $suplimits_{n geqslant 1} sumlimits_{k = 1}^n |c_{n, k}| < +∞$.

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 $limlimits_{n → ∞} sumlimits_{k = 1}^n c_{n, k} = -dfrac{t^2}{2(3 - β)}$. Applying Exercise 3.1.1., $dfrac{S_n}{n^{frac{3 - β}{2}}} Rightarrow cχ$, where $c = dfrac{1}{sqrt{3 - β}}$.