Can we prove a trinomial distribution approaches a Gaussian Consider a random walk with trinomial distribution where $i^{th}$ step is a random variable $X_i$ takes values $-\frac{1}{i}, 0, \frac{1}{i}$ with probability $0.3, 0.4, 0.3$ respectively. (A simpler problem may deal with $X_i$ taking values $-1, 0, +1$ with probability $\frac{1}{3}$ each).
My intuition is that probability distribution for $S_n = \sum_{i \le n} X_i$ is likely a Gaussian. I started by following the template of proof for how the binomial distribution tends to a Gaussian for large n but am having trouble even at the first step when writing equation for $\mathbb{P}(S_n = m)$. Any ideas for how to prove or disprove of this? Thanks!
 A: The $k$-th step has a mean of $\mathbb{E}[X_k] = 0$ and a variance of $\text{Var}[X_k] = \mathbb{E}[X_k^2]-\mathbb{E}[X_k]^2 = \dfrac{0.6}{k^2}$.
Then, $S := \displaystyle\sum_{k = 1}^{\infty}X_k$ has a mean of $\mathbb{E}[S] = 0$, and since the steps $X_k$ are independent, $\text{Var}[S] = \displaystyle\sum_{k = 1}^{\infty}\text{Var}[X_k] = \sum_{k = 1}^{\infty}\dfrac{0.6}{k^2} = \dfrac{\pi^2}{10}$, where we used the well-known sum $\displaystyle\sum_{k = 1}^{\infty}\dfrac{1}{k^2} = \dfrac{\pi^2}{6}$.
The characteristic function of a Gaussian random variable with mean $\mu$ and variance $\sigma^2$ is $e^{i\mu t - \sigma^2t^2/2}$. So if $S$ is Gaussian, its characteristic function $\mathbb{E}[e^{itS}]$ must be $e^{-\pi^2t^2/20}$.
Using the Taylor series expansion $\ln(0.4+0.6\cos x) = -0.3x^2-0.02x^4+O(x^6)$, there exists a $\delta > 0$ such that $\ln(0.4+0.6\cos x) < -0.3x^2$ for all $x \in (0,\delta)$. Then, for any $t \in (0,\delta)$, we have:
\begin{align*} 
\mathbb{E}[e^{itS}] &= \mathbb{E}\left[\exp\left(it\sum_{k = 1}^{\infty}X_k\right)\right]
\\
&= \mathbb{E}\left[\prod_{k = 1}^{\infty}e^{itX_k}\right]
\\
&= \prod_{k = 1}^{\infty}\mathbb{E}\left[e^{itX_k}\right]
\\
&= \prod_{k = 1}^{\infty}\left(0.4+0.6\cos\left(\dfrac{t}{k}\right) \right)
\\
&= \exp\left[\sum_{k = 1}^{\infty}\ln\left(0.4+0.6\cos\left(\dfrac{t}{k}\right) \right)\right]
\\
&< \exp\left[\sum_{k = 1}^{\infty}-0.3 \dfrac{t^2}{k^2}\right]
\\
&= \exp\left[-\dfrac{\pi^2t^2}{20}\right],
\end{align*}
where the inequality step holds by applying the inequality $\ln(0.4+0.6\cos x) < -0.3x^2$ for $x = \dfrac{t}{k}$.
Since $S$ has mean $0$, variance $\dfrac{\pi^2}{10}$, but its characteristic function is not $e^{-\pi^2t^2/20}$, $S$ is not Gaussian.
