By the Borel-Cantelli lemma to prove that
$\Bbb P[\max_{k\le n}x_k<(2+)\sqrt{\log n}$ as $n\to \infty]=1$ with a number $(2+)$ a shade more than 2.
The hint is to use $(1-x)^n>1-nx$.
Let $M_n=\max_{k\ge n}X_k$. Using independence
$$P\{M_n\ge x\}=\prod_{k\le n}P\{X_k\ge x\}=\left[1-\Phi(x)\right]^n$$
For $0<\epsilon<1$ and $m$ s.t. $(1-\epsilon)\sqrt{2\ln n}\ge 1$ for $n\ge m$ we have (using the lower bound for the tail of normal dist. $^{(*)}$)
$$ \sum_{n\ge m}P\{M_n>(1-\epsilon)2\sqrt{\ln n}\}\ge \sum_{n\ge m}\left[1-\frac{n^{-2(1-\epsilon)^2}}{2(1-\epsilon)\sqrt{2\pi\ln n}}\right]^n $$ $$ \ge \sum_{n\ge m}1-n\frac{n^{-2(1-\epsilon)^2}}{2(1-\epsilon)\sqrt{2\pi\ln n}} $$
which is not what I want, since I want this $\sum_{n\ge m}P\{M_n>(1-\epsilon)2\sqrt{\ln n}\}<\infty$, but using the hint, I get the opposite.
Am I using the hint in the wrong way?
Could someone kindly provide some help? Thanks very much!