Suppose that $X_n$ is a sequence of random variables satisfying
$\bullet P(X_n > t\sqrt{n}) < e^{-t^2}$, and
$\bullet X_{m+n} < X_m+X_n$ for all $n,m$.
We want to try and show that $$\limsup_{n\rightarrow \infty} \frac{X_n}{\sqrt{n\log\log(n)}} \leq 1$$ holds almost surely.
My initial attempt was to look at the subsequence $X_{n_k}$ where $n_k = (1+\delta)^k$ for a small $\delta$. Since $$ P(X_{n_k} > (1+\epsilon)\sqrt{n_k\log\log(n_k)}) < e^{-(1+\epsilon)\log\log(n_k)} = \frac{1}{\log(n_k)^{1+\epsilon}} = \left(\frac{1}{k\log(1+\delta)}\right)^{1+\epsilon} $$ Borel-Cantelli implies that $$ \limsup_{k\rightarrow \infty} \frac{X_{n_k}}{\sqrt{n_k\log\log(n_k)}}\leq 1 $$ holds almost surely. I now want to use the subadditivity of the sequence to help me bound the terms $X_n$ for $n_k\leq n < n_{k+1}$. In particular, we have that $$ \frac{X_n}{\sqrt{n\log\log n}} \leq \frac{X_{n_k}}{\sqrt{n_k\log\log n_k}} + \frac{X_{n-n_k}}{\sqrt{n\log\log n}} $$ so that I need to show that $$ \frac{X_{n-n_k}}{\sqrt{n\log\log n}} \rightarrow 0 $$ almost surely. Heuristically, this should be true since $n-n_k \approx \delta (1+\delta)^k$ will be much smaller than $n$, but I'm stuck here. Does anyone have any ideas on how I can bound this $X_{n-n_k}$ term? Is this even a good approach to this problem?