hitting time of Brownian motion

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I'm desperately trying to prove that for a standard BM $B_t$ the stopping time $T:=\inf\{t\geq0: B_t\geq\sqrt{1+t}\}$ is a.s. finite, i.e. $\mathbb{P}[T<\infty]=1$.

I actually tried to play around with $S_t=\sup_{s\in[0,t)}\frac{B_s}{\sqrt{1+s}}$ which should have the half-normal distribution, with the parameters coming from a $\mathcal{N}(0,\frac{s}{1+s})$ distributed r.v. But I cannot see how to prove that $S_t$ must be bigger than one for a certain $t$ a.s.

Thanks in advance!

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The law of the iterated logarithm says that $B_t$ hits the barrier $\sqrt{at\log\log t}$ almost surely and at unbounded times, for every $a\lt2$ (the version of the law for random walks in discrete time is enough). A fortiori, your $T$ is almost surely finite.