Let $d\in\mathbb N$ and $W_1,\ldots,W_d$ be mutually independent, identically distributed and square-integrable real-valued random variables on a probability space $(\Omega,\mathcal A,\operatorname P)$. By the strong law of large numbers, $$\frac1d\sum_{i=1}^dW_i\xrightarrow{d\to\infty}\operatorname E\left[W_1\right]\;\;\;\text{almost surely.}\tag1$$
Now let $$\lambda_d:=\frac1{d^\alpha}$$ for some $\alpha>0$. I would like to conclude $$\liminf_{d\to\infty}\lambda_d^2\sum_{i=1}^dW_i=\begin{cases}-\infty&\text{, if }\alpha<1/2\\0&\text{, if }\alpha>1/2.\end{cases}\tag2$$
It might be crucial to observe $$\lambda_d^2=\frac1{d^{2\alpha-1}}\cdot\frac1d\tag3$$ and $$\lim_{d\to\infty}\frac1{d^{2\alpha-1}}=\begin{cases}\infty&\text{, if }\alpha<1/2\\0&\text{, if }\alpha>1/2.\end{cases}\tag4$$
By $(3)$, $(1)$ and $(4)$ we should at least be able to conclude $$\lambda_d^2\left|\sum_{i=1}^dW_i\right|\le\lambda_d^2\left|\sum_{i=1}^dW_i-\operatorname E\left[W_1\right]\right|+\lambda_d^2\operatorname E\left[W_1\right]\xrightarrow{d\to\infty}0,\tag5$$ if $\alpha>1/2$, and hence the limit inferior in $(2)$ is an actual limit and equal to $0$ as desired.
Assuming that I did no mistake in my reasoning, how can we show the other case $\alpha<1/2$?
EDIT: As the claim seems to be wrong in general, assume that $W_i$ is of the form $$W_i=(\ln f)''(X_i)Z_i^2,$$ where $f\in C^3(\mathbb R)$ with $f>0$, $\int f(x)\:{\rm d}x<\infty$ and $(\ln f)'$ being Lipschitz continuous, $(X_1,\ldots,X_d)$ is a random variable on $(\Omega,\mathcal A,\operatorname P)$ with density $$\mathbb R^d\ni x\mapsto\prod_{i=1}^df(x_i)$$ with respect to the Lebesgue measure on $\mathcal B(\mathbb R^d)$ and $(Z_1,\ldots,Z_d)$ is distributed according to the $d$-dimensional standard normal distribution. It might be worth noting that the assumptions imply that $f(x)\xrightarrow{|x|\to\infty}0$ and $f'(x)\xrightarrow{|x|\to\infty}0$.
The law of the iterated logarithm gives part of the correct answer. Assuming $EW_i=0$ and $EW_i^2=1$, one has $$\liminf_{d\to\infty}\frac {\sum_{i=1}^d W_i}{\sqrt{2d\log\log d}}=-1,$$ from which the critical value of $\alpha=1/4$ not $1/2$ as you have it.
More generally, if $EW_1=\mu$ and $\text{Var}(W_1)=\sigma^2$, one can write $$ \sum_{i=1}^d W_i = d \mu + \sigma \sqrt{2 d \log \log d} \,T_d$$ where with probability $1$, both $\limsup_d \,T_d=1$ and $\liminf_d T_d = -1$. Dividing both sides by $d^{2\alpha}$ gives $$ Q_d = \frac{\sum_{i=1}^d W_i}{d^{2\alpha}} = \frac\mu{d^{2\alpha-1}} + \sigma T_d \frac{\sqrt{2 d \log \log d}}{d^{2\alpha}}.$$ From this one can read off $\liminf Q_d$, depending on the values of $\mu$ and $\alpha$.