Let $B(t)$ be a Standard Brownian Motion.
We need to prove that $\underset{t \to \infty}\limsup \dfrac{B\left(t\right)}{\sqrt{t}\cdot\ln(t)}=0$
I was thinking if we could use the Blumenthal's $0-1$ law along the lines of the proof for $\underset{t \rightarrow \infty}\limsup \dfrac{B(t)}{\sqrt{t}}=\infty$. But I am not sure about how to argue in this case.
Here is perhaps an easier way to argue without the use of Law of Iterated Logarithm or Stochastic integrals.
Let $M_{t}=\sup_{[0,t]}B_{s}$ denote the running max of Brownian Motion and it is a simple consequence of Reflection Principle that $M_{t}\sim |B_{t}|$
The idea is to show that $\frac{M_{s_{k}}}{\sqrt{s_{k}\ln(s_{k})}}<\epsilon$ almost surely for all large $k$ for some increasing sequence $s_{k}$. Then we want to use a sandwich argument by controlling $\frac{B_{t_{k}}}{\sqrt{t_{k}\ln(t_{k})}}$ by means of $\frac{M_{s_{k}}}{\sqrt{s_{k}\ln(s_{k})}}$ by sandwiching properly. This idea can even be used to conclude $\lim\sup\frac{B_{t}}{\sqrt{t\ln(\ln(t))}}\leq 1$ which shows one side of the law of iterated logarithm .
Let us consider a fixed sequence $s_{k}=e^{k}$
Then $P(\frac{M_{e^{k+1}}}{\sqrt{e^{k}\ln(e^{k})}}\geq \epsilon) = P(\frac{M_{e^{k+1}}}{\sqrt{e^{k+1}}}>\epsilon \sqrt{\frac{\ln(e^{k})}{e}}) = P(|X|>\epsilon \sqrt{\frac{\ln(e^{k})}{e}}) $ where $X\sim N(0,1)$ .
and $P(|X|>\sqrt{\frac{\ln(e^{k})}{e}})=2P(X>\sqrt{\frac{\ln(e^{k})}{e}})\leq 2\exp(-\frac{k}{2e}) $ using the gross but useful Gaussian Tail Estimate .
And we also have that $\displaystyle\sum_{k}2\exp(-\frac{k}{2e})<\infty$
Thus by Borel-Cantelli Lemma , only finitely many events $\{\frac{M_{e^{k+1}}}{\sqrt{e^{k}\ln(e^{k})}}\geq \epsilon\}$ occur and hence almost surely $\frac{M_{e^{k+1}}}{\sqrt{e^{k}\ln(e^{k})}}\leq \epsilon$ for all large enough $k$ .
Now let $t_{k}$ be an arbitrary real sequence such that $t_{k}\to\infty$
Then we choose indices $k_{l}$ such that $t_{k}\leq e^{k_{l}+1}$ and $\sqrt{t_{k}\ln(t_{k})}\geq \sqrt{e^{k_{l}}\ln(e^{k_{l}})}$
Then $B_{t_{k}}\leq M_{e^{k_{l}}+1}$ and hence
$$\frac{B_{t_{k}}}{\sqrt{t_{k}\ln(t_{k})}}\leq \frac{M_{e^{k_{l}+1}}}{\sqrt{e^{k_{l}}\ln(e^{k_{l}})}}\leq \epsilon$$ for all large enough $l,k$ .
Thus $\displaystyle\limsup_{k\to\infty}\frac{B_{t_{k}}}{\sqrt{t_{k}\ln(t_{k})}}\leq \epsilon$ .
Now as $t_{k}$ was an arbitrary sequence, you have
$$\lim\sup_{t\to\infty}\frac{B_{t}}{\sqrt{t\ln(t)}}\leq \epsilon$$ and this holds for all $\epsilon$ . Using the fact that $-B_{t}$ is also a Brownian Motion , you also have that $\lim\inf \frac{B_{t}}{\sqrt{t\ln(t)}}\geq -\epsilon$ for all $\epsilon$ . Thus you have your desired conclusion and actually even more , namely , $\lim_{t\to \infty}\frac{B_{t}}{\sqrt{t\ln(t)}}=0$
If you are allowed to use LIL , then the proof is much shorter(actually just a one step proof) .
A point to note is that my choice of the sequence $e^k$ is nothing special and was just made such that the tail bound is nice. I can easily use any $a^k$ for $a>1$.
Let \begin{gather*} A(t)=1+\sqrt{t}\cdot\ln(t\vee 1), \quad t\ge 0, \tag{1}\\ M(t)=\int_0^t\frac{\mathrm{d}B(s)}{A(s)}, \quad t\ge 0. \tag{2} \end{gather*} Then $A$ is positive, continuous and increasing on $\mathbb{R}_+$, $M$ is a continuous martingale on $\mathbb{R}_+$ with \begin{align*} \langle M \rangle_t &=\int_0^t\frac{\mathrm{d}s}{(1+\sqrt{s}\cdot\ln(s\vee 1))^2}\\ &\le \int_0^e\frac{\mathrm{d}s}{(1+\sqrt{s}\cdot\ln(s\vee 1))^2} + \int_e^t\frac{\mathrm{d}s}{s(\ln(s))^2} \\ &\le e + 1.\\ \lim_{t\to\infty} \langle M \rangle_t &< +\infty. \end{align*} Hence \begin{equation*} M(+\infty)=\lim_{t\to\infty}M(t) \quad\text{ exists and $M(+\infty)$ is finite a.s.} \tag{3} \end{equation*}
Furthermore, \begin{equation*} \lim_{t\to\infty}\frac{M(t)}{A(t)}=0. \qquad \text{a.s.} \end{equation*} Now, from (2) and integration by parts, \begin{align*} B(t)&=\int_{0}^{t}A(s)\,\mathrm{d}M(s) =A(t)M(t)-\int_{0}^{t}M(s)\,\mathrm{d}A(s),\\ \frac{B(t)}{A(t)}&=M(t)-\frac{1}{A(t)}\int_{0}^{t}M(s)\,\mathrm{d}A(s). \end{align*} Hence, using (3) get \begin{equation*} \lim_{t\to\infty}\Big[\frac{1}{A(t)}\int_{0}^{t}M(s)\,\mathrm{d}A(s)\Big] =\lim_{t\to\infty}M(t) \tag{4} \end{equation*} and \begin{equation*} \lim_{t\to\infty}\frac{B(t)}{A(t)}=0. \quad \text{a.s.} \end{equation*} Meanwhile, \begin{align*} \lim_{t\to\infty}\frac{\sqrt{t}\ln(t)}{A(t)}=1, \end{align*} Hence, \begin{equation*} \lim_{t\to\infty}\frac{B(t)}{\sqrt{t}\ln(t)}=0.\quad \text{a.s.} \end{equation*}
Add in proof( of (4)): Fixing \begin{align*} \omega_0 & \in \Omega_0 \\ &\stackrel{\triangle}{=}\{\omega: M(+\infty,\omega)=\lim_{t\to\infty}M(t,\omega) \text{ exists and $M(+\infty,\omega)$ is finite }\}, \end{align*} then for each $\varepsilon>0$, there exists $T$ such that \begin{equation*} |M(t,\omega_0)-M(+\infty,\omega_0)|<\varepsilon,\quad \text{as $t>T$}. \end{equation*} Hence \begin{align*} &\Big|\frac{1}{A(t)-A(0) } \int_{0}^{t}M(s,\omega_0)\,\mathrm{d}A(s)-M(+\infty,\omega_0) \Big|\\ &\quad \le \frac{1}{A(t)-A(0)}\int_{0}^{t}|M(s,\omega_0)-M(+\infty,\omega_0)|\,\mathrm{d}A(s)\\ &\quad \le \frac{1}{A(t)-A(0)}\int_{0}^{T}|M(s,\omega_0)-M(+\infty,\omega_0)|\,\mathrm{d}A(s)\\ &\qquad\quad +\frac{1}{A(t)-A(0)} \int_{T}^{t}|M(s,\omega_0)-M(+\infty,\omega_0)|\,\mathrm{d}A(s)\\ &\quad \le \frac{1}{A(t)-A(0)} \int_{0}^{T}|M(s,\omega_0)-M(+\infty,\omega_0)|\,\mathrm{d}A(s) + \varepsilon, \end{align*} and \begin{equation*} \varlimsup_{t\to\infty} \Big|\frac{1}{A(t)-A(0)} \int_{0}^{t}M(s,\omega_0)\, \mathrm{d}A(s)-M(+\infty,\omega_0) \Big|\le \varepsilon. \end{equation*} Now letting $\varepsilon\downarrow 0$ get \begin{gather*} \varlimsup_{t\to\infty} \Big|\frac{1}{A(t)-A(0)}\int_{0}^{t}M(s,\omega_0)\, \mathrm{d}A(s)-M(+\infty,\omega_0) \Big|=0. \end{gather*} This also means \begin{equation*} \lim_{t\to\infty} \frac{1}{A(t)}\int_{0}^{t}M(s,\omega_0)\,\mathrm{d}A(s) = M(+\infty,\omega_0), \end{equation*} and \begin{equation*} \Omega_0\subset \Big\{\omega:\lim_{t\to\infty} \frac{1}{A(t)}\int_{0}^{t}M(s,\omega)\,\mathrm{d}A(s) = M(+\infty,\omega) \Big\}. \end{equation*}
