Proving convergence in distribution using Levy's continuity theorem I'm trying to solve the following question - parts (a) and (b) appear to be very similar in structure but I cannot solve part (b):


My attempt:
For part (a), we apply Levy's continuity theorem.  Fix $u \in \mathbb{R}$ and note $$E\left(\exp\left(i\frac{uY_t}{\sigma_M\sqrt{t}} \right)\right) = E\left(\sum_{n=0}^\infty \mathbf{1}(N_t = n)\exp\left(i\frac{u \sum_{k=1}^n X_M(k)}{\sigma_M\sqrt{t}} \right)\right) \\ = \sum_{n=0}^\infty \frac{e^{-t}t^n}{n!}E\left(\exp\left(i\frac{u \sum_{k=1}^n X_M(k)}{\sigma_M\sqrt{t}} \right)\right) \\ = e^{-t}\sum_{n=0}^\infty \frac{1}{n!}\left(t E\left(\exp\left(i\frac{u  X_M(1)}{\sigma_M\sqrt{t}} \right)\right)\right)^n \\ = \exp \left(-t + t E\left(\exp\left(i\frac{u  X_M(1)}{\sigma_M\sqrt{t}} \right)\right)\right)$$
by the independence of $N_t$ and the $X_M(k)$ and applying dominated convergence to interchange the sum and expectation for the second equality and by the iid property of the $X_M(k)$ for the third.  We will deal only with the exponent for now, and for shorthand we define $Z \equiv X_M(1)$:
$$-t + tE\left(\exp\left(i\frac{u  Z}{\sigma_M\sqrt{t}} \right)\right) = -t + tE\left(\sum_{j=1}^\infty \frac{i^j u^j  Z^j}{\sigma_M^j t^{j/2} j!} \right) \\ = -t + t\left(1 + 0 + \frac{i^2E(Y^2)u^2}{2\sigma_M^2 t} + \sum_{j=3}^\infty \frac{i^j u^j  E(Z^j)}{\sigma_M^j t^{j/2} j!} \right) $$ where we again apply DCT and note that by the symmetry of the distribution for $Z$ that it's expectation is 0.
$$= -\frac{u^2}{2} + \frac{1}{\sqrt{t}} \sum_{j=3}^\infty \frac{c^j E(Z^j)}{j!} \cdot \frac{1}{t^{(j-3)/2}} \quad \quad \quad \textbf{(L)}\\ \xrightarrow{t \rightarrow \infty} -\frac{u^2}{2}$$
where $c = \frac{i u}{\sigma_M}$.  For every $t \ge 1$ and the sum above has bounded modulus (by $\exp(|c|M)$ for example), thus justifying the convergence of the characteristic function to that of a $N(0,1)$ and we may conclude part (a).

For part (b), I have been trying to do the same thing, which will obviously require the computation of $\sigma_M$ since we didn't use that in part (a).  It is trivially shown that (for brevity put $\Delta \equiv \arctan(M) - \arctan(-M)$)  $$\sigma_{M(t)} = \sqrt{E(X_{M(t)}(1)^2)} = \sqrt{\frac{2M - \Delta}{\pi\Delta}}$$
I believe that the convergence after line (L) can hold if and only if $$\sum_{j=3}^\infty \frac{c^j E(Z^j)}{j!} \cdot \frac{1}{t^{(j-3)/2}} \xrightarrow{t \rightarrow \infty} 0$$  I have tried rewriting the modulus of the sum to include all the information about $\sigma_{M(t)}$, i.e. as being equal to $$\lvert\sum_{j=3}^\infty \frac{c^j E(Z^j)}{j!} \cdot \frac{1}{t^{(j-3)/2}}\rvert \leq \sum_{j=3}^\infty \frac{u^j}{j!} \left(\frac{M(t)^2\pi \Delta}{2M-\Delta}\right)^{j/2} \cdot \frac{1}{t^{(j-3)/2}} $$  I have no idea how to make this conclusion from here though.  Please help if you can - I've wasted a stupid amount of time on this.
 A: By noting that there exists a constant $C > 0$ for which
$$ \left| e^{ix} - \left( 1 + ix - \frac{x^2}{2} \right) \right| \leq Cx^3 \tag{*} $$
holds for all $x \in \mathbb{R}$, we have
\begin{align*}
&\left| t \mathbb{E}\left[\exp\left(\frac{iuX_M}{\sigma_M\sqrt{t}}\right)\right] - \left(t - \frac{u^2}{2} \right) \right| \\
&\leq \frac{C u^3}{\sigma_M^3 \sqrt{t}} \mathbb{E}\bigl[|X_M|^3\bigr]
\leq \frac{C u^3}{\sigma_M^3 \sqrt{t}} \mathbb{E}\bigl[M X_M^2\bigr]
\leq \frac{C M u^3}{\sigma_M \sqrt{t}}.
\end{align*}
Now by noting that
$$ \sigma_M \sim \frac{M}{\sqrt{3}} \quad\text{as}\quad M\to 0^+ \qquad\text{and}\qquad \sigma\sim\sqrt{\frac{2}{\pi}M} \quad\text{as}\quad M\to\infty,$$
we can further bound the difference as
$$ \left| t \mathbb{E}\left[\exp\left(\frac{iuX_M}{\sigma_M\sqrt{t}}\right)\right] - \left(t - \frac{u^2}{2} \right) \right| \leq 
C_2u^3 \frac{\max\{1,\sqrt{M}\}}{\sqrt{t}} $$
for an absolute constant $C_2 > 0$. Since this bound converges to $0$ as $t \to \infty$ by the assumption on $M$, the desired conclusion follows.

Addendum.

*

*I believe that $\pi$ in the denominator of $\text{(5)}$ is a typo. The correct formula would be
$$ f_{X_M}(x) = \frac{1}{2\arctan(M)} \frac{\mathbf{1}_{\{|x| \leq M\}}}{1+x^2}. $$


*Validity of $\text{(*)}$ critically hinges on the restriction $x \in \mathbb{R}$, and so, it cannot be directly obtained from the power series expansion. However, this can be proved by using an explicit formula for the remainder term in the Taylor approximation. For instance, we may utilize
$$ e^{ix} = 1 + ix - \frac{x^2}{2} + \frac{x^3}{2i} \int_{0}^{1} (1-s)^2 e^{ixs} \, \mathrm{d}s, $$
thus proving $\text{(*)}$ with $C = \frac{1}{6}$.
