Does the distribution of T converge to a normal distribution despite not satisfying the Lyapunov condition?

Question

For a sequence of independent and identically distributed (i.i.d) random variables $S_i$ where $i = 1, 2, \dots, n$. Each $S_i$ can take values (-1, 1) with probability $\left(\frac{1}{2}, \frac{1}{2}\right)$.

Now, consider another sequence of random variables defined by $$ T_i = \frac{S_i}{i} $$ for $i = 1, 2, \dots, n$.

Define T: $$ T = T_1 + T_2 + \dots + T_n $$

Question: Will the distribution of $ T $ converge to a normal distribution when $n$ approaches infinity?

While each $ T_i $ is independent, they are not identically distributed. We've observed that this sequence does not satisfy the Lyapunov condition for the Central Limit Theorem. However, our numerical simulations seem to suggest that the distribution of $ T $ does converge to a normal distribution. We are uncertain about the conclusions drawn from our simulations. Perhaps our simulations lack sufficient accuracy? Any insights would be appreciated.

Answer 1

Empirically, no: the convergence is instead to a platykurtic distribution with mean $0$ and varaince $\frac{\pi^2}{6}$, clearly not normal. (For finite $n$ the distribution is discrete and with variance marginally below $\frac{\pi^2}{6}$, but the limiting distribution is continuous.)

This should not be a surprise: the distribution of the sum will have a finite positive limit of the variance without further scaling as $n$ increases (unlike the central limit theorem) so peculiarities with small $i$ never get lost in the sum or average. If it had been normal then you might have guessed it would also have been normal without the $S_1$ term, but adding back the $S_1$ term would turn it into a non-normal mixture with a flattened peak. $\frac{S_2}{2}$ would have a similar peak-flattening effect though smaller, and so on.

Here is an example using R of the empirical density from a million samples each for $n=100$ (blue) and $n=1000$ (red), and the corresponding normal density in black. Note that the red line largely overlaps the blue line, even allowing for simulation noise. The black normal density is very different in shape.

T <- function(n){
  S <- sample(c(1,-1), n, replace=TRUE)
  return(sum(S / (1:n)))
}

set.seed(2023)
curve(dnorm(x, 0, sqrt(pi^2/6)), from=-5, to=5) 
lines(density(replicate(10^6, T(100))), col="blue")
lines(density(replicate(10^6, T(1000))), col="red")