Consider a Bernoulli random variable denoted $B_i$, representing a coinflip.
It is clear that a Random variable:
$$X_n:=\sum_{i=1}^{i=n}B_i$$
is a Binomial random variable: $X_n \sim Bin(n,p)$. Using CLT, we have that:
\begin{equation} \label{1} \frac{X_n-np}{\sqrt{np(1-p)}}\to^{D}N(0,1) \tag{1} \end{equation}
It is well known that a Normal distribution might be used as an approximation for the Binomial, and in fact I have come across the following statement (again, using CLT):
\begin{equation} \label{2} X_n \to^{D}N\left(np, np(1-p)\right) \tag{2} \end{equation}
Question: I wonder if \ref{2} above is rigorous? It is obvious that as $n\to\infty$, both the mean $np$ and the variance $np(1-p)$ diverge. This also makes sense logically, because as the number of coinflips tends to infinity, one would of course expect that the 'expected value' of successes $np$ will also tend to infinity.
So can one use \ref{2} with mathematical rigor, or is only \ref{1} mathematically rigorous?
Depends on what you mean by rigorous.
Saying a succession tends to a certain limit as $n \to \infty$ means in particular thta, for $n$ large enough, the succession is close to the limit. Thus if $Z \sim \mathcal{N}(0,1)$, for $n$ large
$$ \mathbb{P}\bigg( \frac{\bar{X}_n - \mu}{\sigma /\sqrt{n}} < t \bigg) \approx \mathbb{P}(Z < t) $$
This can be written as
$$ \mathbb{P}(\bar{X}_n < k) \approx \mathbb{P}\bigg( \frac{\sigma}{\sqrt{n}}Z + \mu < k\bigg) = \mathbb{P}(Y < k), \quad Y\sim \mathcal{N}\bigg(\mu, \frac{\sigma^2}{n}\bigg) $$
Alternatively again we have
$$ \mathbb{P}\bigg( \sum_{i=1}^{n}X_i < k \bigg) \approx \mathbb{P}(Y < k), \quad Y\sim \mathcal{N}\bigg(n\mu, n\sigma^2\bigg) $$
This last form is the one in (2) if applied to a binomal rv.
As a side note: the binomial is symmetric for $p=1/2$ and the closer it is to $p \in \{0,1\}$ the more asymmetric it becomes. For this reason, usually, we apply the normal approximation when $np > 5$ and $n(1-p) > 5$