Sum of $n$ Poisson random variables with parameter $\frac 1 n$ I have been working through the exercises of a textbook and stumbled upon the question as follows:

A skeptic gives the following argument to show that there must be a ﬂaw in
the central limit theorem: “We know that the sum of independent Poisson random variables follows a Poisson distribution with a parameter that is the sum
of the parameters of the summands. In particular, if $n$ independent Poisson
random variables, each with parameter $n^{−1}$, are summed, the sum has a Poisson distribution with parameter $1$. The central limit theorem says that as $n$
approaches inﬁnity, the distribution of the sum tends to a normal distribution,
but the Poisson with parameter $1$ is not the normal.” What do you think of this
argument?

My first intuition is that he is wrong, but I do not quite know how to put it into words. I believe that the flaw is in assuming that the sum of $n$ terms as $n \rightarrow \infty$ will still result in a Poisson random variable with parameter $1$. But I do not quite know how to show whether this is true.
Any hints would be greatly appreciated!

This 'problem' arises because the number $n$, whose limit is taken in the central limit theorem (CLT), is coupled to properties of the terms in the sum.
Let $$X_i \sim Pois(m^{-1})$$ and consider the standardized sum $$S_{m,n} = \frac{-n/m  + \sum_{i=1}^n X_i}{\sqrt{n/m}} = \frac{\sum_{i=1}^n X_i-1/m}{\sqrt{n/m}} $$
The following limit will be the correct expression of the CLT
$$\lim_{n \to \infty} S_{m,n} \to N(0,1)$$
This is not the same as
$$\lim_{(m,n) \to (\infty,\infty)} S_{m,n} \to N(0,1)$$
In some cases, this other limit may be true as well, but it will depend on how the parameters $m$ and $n$ go to infinity.
It is not correct, but why does it not work?
Intuitive
The above part is the same point as Dilip Sarwate has made in his answer. The central limit theorem does not relate to changing the parameters $m$ and $n$ simultaneously.
Still, while the CLT has technically not been applied correctly, we do feel in an intuitive sense that a sum of many little variables should approach a Gaussian distribution. One might still wonder what is happening and why this occurs. Why doesn't the CLT work in this way?
If we would do the same with an exponential distributed variable $X_i \sim exp(n)$ then the standardized sum of $n$ of those variables would converge to a Gaussian distributed variable. Why is this not the case for the Poisson distributed variable?
Summing makes the shape of distribution less important
The reason is closely related to the Poisson distribution being infinitly divisible. This might be seen as similar to the central limit theorem. Some distributions can be expressed as a sum of i.i.d distributed variables with the same distribution family. When you sum the variables you get a variable from the same family back. The Gaussian distribution is a well-known example. The Levy distribution and Cauchy distribution have the same property (and they can also arise as a limit distribution when summing many variables).
The 'trick' of the central limit theorem is that summing variables (while shifting and scaling to keep the same mean and variance) will make the specific shape of the distribution less dominant in defining the end result. In the proof of the CLT by means of the characteristic functions, you can see this as only the first terms of a Taylor expansion for the characteristic function counting, while the rest becomes smaller.
In terms of cumulants
In terms of cumulants, it is more easily seen. For cumulants we can use the following properties for the $k$-th cumulant
$$\begin{array}{}
\kappa_{1}(c+X) &=& \kappa_{1}(X) + c \\
\kappa_{k}(c+X) &=& \kappa_{n}(k) \quad \quad \text{for $k \geq 2$}  \\
\kappa_{k}(cX) &=& c^k \kappa_{k}(X) \\
\kappa_{k}(X+Y) &=& \kappa_{k}(X)+\kappa_{k}(Y) \\
\end{array}$$
such that
$$\begin{array}{}
\kappa_{1}(S_{m,n}) &=& \mu_S = 0\\
\kappa_{2}(S_{m,n}) &=& \mu_S^2 + \sigma_S^2 = 1\\
\kappa_{k}(S_{m,n}) &=& n \left( \sqrt{\frac{1}{n\sigma_{X_{m}}^2}} \right)^k \kappa_k(X_{m}) \quad  \quad \text{for $k  \geq 3$} 
\end{array}$$
With $X_m$ we denote the variable $X$ with parameter $m$.
For the Poisson distribution the cumulant generating function (the log of the moment generating function) is:
$$g(t) = \log M(t) = \frac{e^t-1}{m}$$
The $k$-th cumulants are the $k$-th derivatives in the point $t=0$
$$\kappa_k = g(t)^{(m)} =   \frac{1}{m}$$
So when we increase the number of variables $n$ then we decrease the higher order cumulants and the shape of the distribution becomes less important and the result is more like a Gaussian distribution. The reason why it does not happen in the case of the question is that changing the parameter $m$ makes the shape more pronounced and counters the effect of the central limit theorem.

Generalizing
In terms of the cumulants, we can see a generalization of the situation. Consider that the increase of the number of variables in the sum $n$ is associated with a change of the parameters of the distribution of the variable in the sum. If this is such that the higher-order cumulants, of the standardized variable, increase with a rate of at least $ n^{k/2-2}$
$$ \liminf_{n \to \infty} \frac{\kappa_k(X_n/\sigma_{X_n})}{n^{k/2-2}} = \liminf_{n \to \infty} \frac{\kappa_k(X_n)}{\kappa_2(X_n)^{k/2}} \frac{1}{ n^{k/2-2}} > 0$$
then the higher-order cumulants of the summation won't decrease in the limit and the distribution does not approach a Gaussian distribution.
Example 1
Let's try to do the same with a Bernoulli distribution. The ratio of the second and third cumulant is
$$\frac{\kappa_3(X_n)}{\kappa_2(X_n)^{3/2}} = \frac{(1-2p)}{ \sqrt{p(1-p)}}$$
and if we put this equal to $\sqrt{n}$ or if we use $n = \lceil (1-2p)/p(1-p) \rceil$ then the summation should not approach a Gaussian distribution.
What we will get is a sum of $n$ Bernoulli variables with approximately $p \approx 1/n$ and this will be equal to a Poisson distribution.
Example 2
Let's try it with the exponential distribution.
$$\frac{\kappa_3(X_n)}{\kappa_2(X_n)^{3/2}} = \frac{2\lambda^{-3}}{(\lambda^{-2})^{3/2}} = 2$$
This time there is no dependence on the parameter and no matter how we adjust $\lambda$ as a function of $n$, the normalized sum will always approach a Gaussian distribution.
Example 3
Let's try it with the gamma distribution with fixed $\theta = 1$.
$$\frac{\kappa_3(X_n)}{\kappa_2(X_n)^{3/2}} = \frac{k}{(3k)^{3/2}} = \frac{1}{\sqrt{27 k}}$$
So if we let $k = 1/n$ then we should get a distribution that approaches something that is not a Gaussian distribution.
Knowing the properties of the gamma distribution (sums of gamma with the same scale are gamma as well) this approaches a gamma distribution with parameter $k=1$.
