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!

There are a lot of great answers already but I want to add a perspective interpreting this in light of Lindeberg's CLT. I think this is a natural way to look at it because we have a collection of random variables $\{X_{nj} : n= 1,2\dots\text{ and }j = 1,\dots,n\}$ with all independent and $X_{nj} \sim \text{Pois}(n^{-1})$. This is a triangular array and, letting $S_n = \sum_{j\leq n} X_{nj}$ be the sum of the $n$th row, Lindeberg's CLT gives us a sufficient condition for when
$$
\frac{S_n - \text E[S_n]}{\sqrt{\text{Var}[S_n]}} \stackrel{\text d}\to \mathcal N(0,1).
$$
Unlike the simplest form of the CLT, Lindeberg's CLT allows for non-identically distributed RVs so the changing distribution of the $X_{nj}$ alone won't be what prevents convergence to a Gaussian here.

The theorem is as follows (adapted from Jun Shao's Mathematical Statistics Theorem 1.15 2nd ed.):

Consider a triangular array of random variables $\{X_{nj} :  n=
 1,2,\dots \text{ and }j = 1, \dots, n\}$ where all are independent and
let $S_n = \sum_{j\leq n} X_{nj}$ be the sum across each row of the
array.  The Lindeberg CLT states that if $0 < \sigma_n^2 :=
 \text{Var}(S_n) < \infty$ for $n=1,2,3,\dots$ and
if$\newcommand{\e}{\varepsilon}$ Lindeberg's condition holds, which
is that $$ \sigma^{-2}_n\sum_{j=1}^n \text E[(X_{nj} - \text
 EX_{nj})^2\mathbf 1_{|X_{nj} - \text EX_{nj}| > \e\sigma_n}] \to 0 \tag{1}$$
for all $\e > 0$ as $n\to\infty$, then $$ \sigma_n^{-1}\sum_{j=1}^n
 (X_{nj} - \text EX_{nj}) = \frac{S_n - \text
 E[S_n]}{\sqrt{\text{Var}[S_n]}} \stackrel{\text d}\to \mathcal N(0,
 1). $$

Lindeberg's condition means that the variance of the row sum dominates the sum of the tail variances of each element as we advance down the triangular array, and in particular the tails relative to the row sum variance. Intuitively, this will fail to hold if any of the individual $X_{nj}$ are heavy tailed enough to contribute a lot relative to the row sum variance.

In our case we know that we have $S_n\sim\text{Pois}(1)$ for all $n$ so $\sigma^2_n = 1 < \infty$. Lindeberg's condition in this case is
$$
n \operatorname E[(X_{n1} - 1/n)^2\mathbf 1_{|X_{n1} - 1/n|> \e}] \stackrel ?\to 0
$$
for all $\e > 0$.
If the conclusion were true then we'd have $S_n - \text E[S_n] \stackrel{\text d}\to \mathcal N(0,1)$ which is certainly not true as $S_n - \text E[S_n] \sim \text{Pois}(1)-1$ and doesn't change with $n$. This is a nondegenerate limiting distribution but it's not Gaussian so there must be something wrong about the condition.
Let $\e > 0$. I'll drop the "$1$" from the subscript and just use $X_n$, so $X_n := X_{n1}$. The condition $|X_n - 1/n|>\e$ means that we only want to consider values $x$ of $X_n$ where $x > 1/n+\e$ or $x < 1/n - \e$; in other words, at least $\e$ away from $1/n$. We are taking $n\to\infty$ so we only need to consider $n$ when sufficiently large. This needs to hold for any $\e$, but the bigger $\e$ is then the fewer $x$ we consider, so the worst case will be $\e$ small.
$\e$ is fixed and eventually $1/n < \e$ which means $x < 1/n-\e < 0$ and that's a place with zero probability for the Poisson, so we only need to consider $x>1/n+\e$ which is just $x > 0$ eventually. Lindeberg's condition is therefore about the limit of
$$
n \operatorname E[(X_{n} - 1/n)^2\mathbf 1_{|X_{n} - 1/n|> \e}] = n \sum_{x > 0} (x - 1/n)^2 P(X_n = x) \\
= n \left(\operatorname E[(X_n-1/n)^2] - \frac 1{n^2}P(X_n = 0)\right) \\
= n\left(\frac 1n  - \frac{1}{n^2}e^{-1/n}\right) \\
= 1  - \frac 1n e^{-1/n} \to 1.
$$
Thus Lindeberg's condition fails to hold. The problem is that the mean of the $X_{nj}$ moves to zero as they do too, so the tail variances are too large relative to the row sum variance which is constant. We don't lose the effects of the individual summands in the way that we need for the CLT to hold.
In general Lindeberg's condition is not necessary, i.e. it can fail to hold yet the CLT result still applies, but since we know the limiting distribution is not Gaussian I think we can still gain insight from the failure of the condition.
