In the construction of Brownian motion from scaled symmetric random walks, we consider the process
$$W^{(n)}(t) = \frac{1}{\sqrt{n}} M_{nt}$$
where $M_{nt}$ is the sum of $nt$ independent coin tosses $\{\pm 1\}$ when $nt$ is an integer (otherwise a linear interpolation is done). Central Limit Theorem will tell us that $W^{(n)}(t)$ converges (in distribution) to a normal random variable, for each $t>0$.
Now, only as a fun exercise, I want to analyze what happens if we change this scaling factor from $\frac{1}{\sqrt{n}}$ to say $f(n) = \frac{1}{n^c}$, where $c \in \mathbb{R}$. My intuition is that if $c>\frac12$ then the limit should be the constant zero process, otherwise if $c < \frac12$ then it should be a divergent process. To do so, I will make use of the fact that convergence in distribution is equivalent to convergence of their characteristic functions.
So now my new scaled symmetric random walk is $W^{(n)}(t) = f(n) M_{nt}$. Let us compute characteristic function $\phi_n(u)$ of $W^{(n)}(t)$
\begin{align*} \phi_n(u) &= \mathbb{E}\left(e^{iuW^{(n)}(t)}\right) = \mathbb{E}\left(e^{iuf(n)M_{nt}}\right) \\ &= \mathbb{E}\left(\prod_{i=1}^{nt} e^{iuf(n)X_i}\right) = \left(\mathbb{E} e^{iuf(n)X_i}\right)^{nt} \\ &= \left(\frac{1}{2} (e^{iuf(n)} + e^{-iuf(n)})\right)^{nt} \\ &= \left(\cos(uf(n))\right)^{nt} \\ \end{align*}
(I think there are some subtleties to be taken care of in the above work, namely for a fixed $t$, $nt$ will not be integer for all $n$, so perhaps some more justification is needed, but I will turn a blind eye)
Now $\phi_n(u) = \left(\cos(\frac{u}{n^c})\right)^{nt} = \exp \left(nt \log \cos(\frac{u}{n^c}) \right)$. Setting $x = \frac{1}{n}$, we get
$$\lim_{n \to \infty} \phi_n(u) = \lim_{x \to 0^+} \exp \left(ut \frac{\log \cos ux^c}{ux}\right)$$
If $c>0$, applying L'Hôpital's rule gives:
$$\lim_{x \to 0^+} \frac{\log \cos ux^c}{ux} = \lim_{x \to 0^+} - cx^{c-1} \tan ux^c= -uc \left( \lim_{x \to 0^+}\frac{\tan ux^c}{ux^{1-c}} \right)$$
For verification, notice that when $c=\frac12$, the above expression evaluates to $-\frac{u}{2}$, which gives $\phi_n(u) \xrightarrow{n \to \infty} e^{-\frac{u^2t}{2}}$ as desired (characteristic function of $N(0,t)$).
Case 1: $c = \frac12 + \epsilon$ (where $\epsilon > 0$)
$$\lim_{x \to 0^+}\frac{\tan ux^c}{ux^{1-c}} = \lim_{x \to 0^+}\frac{\tan ux^{\frac12 + \epsilon}}{ux^{\frac12 + \epsilon}} x^{2\epsilon} = 0$$
Hence $\phi_n(u) \xrightarrow{n \to \infty} 1$ as we had expected. So the limiting process is the constant zero (almost surely) process.
Case 2: $c = \frac12 - \epsilon$ (where $0 < \epsilon < \frac12$)
$$\lim_{x \to 0^+}\frac{\tan ux^c}{ux^{1-c}} = \lim_{x \to 0^+}\frac{\tan ux^{\frac12 - \epsilon}}{ux^{\frac12 - \epsilon}} \frac{1}{x^{2\epsilon}} = \infty$$
The characteristic functions $\phi_n(u) \to \infty$, so what conclusion follows? I am feeling a little lost here.
Case 3: $c \leq 0$. Let us replace $c$ with $-c$ (for convenience) and note that
$$\lim_{n \to \infty} \phi_n(u) = \lim_{n \to \infty} \left(\cos un^c\right)^{nt}$$
In particular notice that, if $f(n) = n$, then $\phi_n(u) = (\cos un)^{nt}$ doesn't have a limit because $\phi_n(\pi) = (-1)^{n^2t}$ doesn't have a limit as $n \to \infty$ (at least for $t=1$)
So the scaled symmetric random walk doesn't always converge/diverge. The limiting process might not even exist. At first glance I don't find that very surprising because in the above random walk ($f(n)=n$), the scaling factor tends to $\infty$ so intuitively it seems like "the limiting process is fluctuating between $\pm \infty$ which is strange" and hence the limiting process doesn't make sense.
Finally, can someone help me complete my analysis by helping me with that limit in case 3? Seems like the limit doesn't exist whenever $c \leq 0$. Not sure how to prove that. Will really appreciate some help!
EDIT: In case 3, we make three more subcases:
Case 3a: $c=0$ then $\phi_n(\pi) = (\cos \pi)^{nt} = (-1)^{nt}$ which doesn't converge for $t=1$
Case 3b: $c \in \mathbb{Q}$ then let $c = \frac{p}{q}$ and consider the subsequence $\phi_{n^q}(\pi) = (-1)^{nt}$ which again doesn't converge for $t=1$ and hence the original sequence doesn't converge.
Case 3c: $c \in \mathbb{R} \setminus \mathbb{Q}$. I am not sure of this case but perhaps some argument like following can be completed to show that $\phi_n(\pi)$ doesn't converge.
Fix a sequence $r_k$ of rationals which converges to $c$. Let us denote by $a_{n} = (\cos \pi n^c)^{n}$ and $b^k_n = (\cos \pi n^{r_k})^{n}$.
We know two things: $$\lim_{n \to \infty} b^k_n \text{ doesn't exist}$$ $$\lim_{k \to \infty} b^k_n = a_n$$
In particular, if somehow I can justify that limit can be interchanged in the following, then I am done: $\lim_{n \to \infty} \lim_{k \to \infty} b^k_n$