= Solution
Use the <Donsker invariance principle>: for <independent and identically distributed random variables> of mean zero and variance one, the linearly interpolated processes $W_n$ with $W_n(k/n)=S_k/\sqrt n$ converge in distribution to standard <Brownian motion> in $C[0,1]$ with the uniform topology. No moment beyond the finite second moment is required.
The maximum functional $F(f)=\max_{0\leq t\leq1}f(t)$ is continuous, since $|F(f)-F(g)|\leq\|f-g\|_\infty$. A linear function on each interpolation interval has its maximum at an endpoint, so $F(W_n)=M_n/\sqrt n$. The <continuous mapping theorem> therefore gives
$$
\frac{M_n}{\sqrt n}\xrightarrow{d}\max_{0\leq t\leq1}B_t.
$$
By the <Brownian reflection principle>, the limiting <random variable> has the distribution of $|Z|$, $Z\sim N(0,1)$. Its <distribution function> is continuous and has no atom at any $x\geq0$, so convergence in distribution permits passage to these tail probabilities. Hence
$$
\boxed{\lim_{n\to\infty}\mathbb P(M_n\geq x\sqrt n)
=2(1-\Phi(x))=\frac2{\sqrt{2\pi}}\int_x^\infty e^{-y^2/2}\,dy\quad(x\geq0).}
$$
At $x=0$ the left side is exactly one for every $n$, because $S_0=0$ is included in the maximum, and the right side is also one.
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