Find limit distribution of given random variable I'm struggling doing the following exercise formally:

Let $(X_i)_{i\in \mathbb N}$ i.i.d. random variables with $\mathbb E[X_1] = 0$ and $\mathbb Var(X_1) = 1. $Find a random variable $Z$ such that $$S_n:=n^{-3/2} \sum_{k=1}^n kX_k \longrightarrow Z$$
  in distribution.  

Solution: I think I've got the right solution, but I can't formally justify taking the limits.
Since we have mean $0$ and variance $1$, the Taylor Expansion of $$\varphi_{X_1}(t) = 1-t^2/2+o(t^2) \quad \text{as }t\to0.$$
Now we get $$\varphi_{n^{-3/2} \sum_{k=1}^n kX_k}(t) = \prod_{k=1}^n\varphi_{X_k}(n^{-3/2}kt) = \prod_{k=1}^n\left(1-\frac{k^2t^2}{2n^3} + o\left(\frac{k^2t^2}{n³}\right)\right).$$
Now the expression inside the "$o$" is what I don't like and I don't know how to handle. If we take the logarithm and use $\log(x) = x + o(x)$ we obtain $$\varphi_{S_n}(t) = -\sum_{k=1}^n \left(\frac{k^2t^2}{2n^3} + o\left(\frac{k^2t^2}{n³}\right)\right).$$
The first expression in the sum I can calculate by using $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and get $e^{-t^2/6}$ as the pointwise limit. But how to prove that $$\sum_{k=1}^n o\left(\frac{k^2t^2}{n³}\right)$$
goes to zero, if that is even a well defined expression? What else could I do to avoid using the $o$ notation without the calculations getting too messy? Any help appreciated!
 A: There is the following elementary result which is not difficult to prove (e.g. by induction):

Let $(a_k)_{k=1,\ldots,n}$ and $(b_k)_{k=1,\ldots,n}$ be complex numbers such that $|a_k| \leq 1$ and $|b_k| \leq 1$ for all $k \in \{1,\ldots,n\}$. Then $$\left| \prod_{k=1}^n a_k - \prod_{k=1}^n b_k \right| \leq \sum_{k=1}^n |a_k-b_k|. \tag{1}$$

Let's use this lemma to write down a formal proof. By Taylor's formula, there exists a locally bounded mapping $\varphi: \mathbb{R} \to (0,\infty)$, $\lim_{t \to 0} \varphi(t)=0$, such that
$$\varphi_{X_1}(t) = 1- \frac{t^2}{2} + t^2 \varphi(t) \tag{2}$$
Using
$$\varphi_{S_n}(t) = \prod_{k=1}^n \varphi_{X_1}(n^{-3/2} kt)$$
and the above Lemma we find that
$$\begin{align*} \left| \varphi_{S_n}(t) - \prod_{k=1}^n \left(1- \frac{k^2 t^2}{2n^3} \right) \right| &\leq \sum_{k=1}^n \left|\varphi_{X_1}(n^{-3/2} kt)- \left(1- \frac{k^2 t^2}{2n^3} \right) \right| \\ &\stackrel{\text{(2)}}{\leq} \sum_{k=1}^n \left( \frac{k^2 t^2}{n^3} \varphi \left[ \frac{k^2 t^2}{n^3} \right] \right) \tag{3} \\ &\leq t^2 \sup_{|s| \leq t^2/n} \varphi(s). \end{align*}$$
As $\varphi(r) \to 0$ as $r \to 0$, the right-hand side converges to $0$ as $n \to \infty$. Thus we have shown that
$$\lim_{n \to \infty} \left| \varphi_{S_n}(t) - \prod_{k=1}^n \left(1- \frac{k^2 t^2}{2n^3} \right) \right| = 0. \tag{4}$$
Since
$$\log \left( \prod_{k=1}^n \left[1- \frac{k^2 t^2}{2n^3} \right] \right) = \sum_{k=1}^n \log \left(1- \frac{k^2 t^2}{2n^3} \right)$$
and, by Taylor's formula,
$$|\log(1+x)-x| \leq |x| \tilde{\varphi}(x)$$
for some mapping $\tilde{\varphi}$ such that $\lim_{x \to 0} \tilde{\varphi}(x) =0$, we otain that
$$\begin{align*} \left| \log \left( \prod_{k=1}^n \left[1- \frac{k^2 t^2}{2n^3} \right] \right)  - \sum_{k=1}^n \frac{-k^2t^2}{2n^3} \right| &\leq \sum_{k=1}^n \frac{k^2 t^2}{2n^3} \tilde{\varphi} \left( \frac{k^2 t^2}{2n^3} \right) \end{align*} $$
Exactly as in $(3)$ we find that the right-hand side converges to $0$ as $n \to \infty$. Thus,
$$\lim_{n \to \infty} \prod_{k=1}^n \left(1- \frac{k^2 t^2}{2n^3} \right) = \lim_{n \to \infty} \exp \left(- \sum_{k=1}^n \frac{k^2 t^2}{2n^3} \right) = \exp \left(- \frac{t^2}{6} \right).$$
Combining this with $(4)$ we conclude that
$$\lim_{n \to \infty} \varphi_{S_n}(t) = \exp \left( - \frac{t^2}{6} \right)$$
for all $t \in \mathbb{R}$.
