= Solution
{c}
For $\lambda>1$, use $x/\lambda\geq1$ on the tail:
$$
\int_\lambda^\infty e^{-x^2/2}\,dx
\leq\frac1\lambda\int_\lambda^\infty xe^{-x^2/2}\,dx
=\frac{e^{-\lambda^2/2}}\lambda
\leq e^{-\lambda^2/2}.
$$
This proves the requested <Gaussian tail bound>.
\b[The subsequent printed maximum inequality is false as stated.] For $n=1$ its integrand never reaches one, and its right-hand side is
$$
\frac2{\sqrt{2\pi}}\int_0^\infty e^{-x^2/2}\,dx=1.
$$
But for two independent absolute standard normal variables, using their common distribution function $F(x)=2\Phi(x)-1$ and the <standard normal density> $\varphi$,
$$
\mathbb E M_1
=4\int_0^\infty x(2\Phi(x)-1)\varphi(x)\,dx
=8\int_0^\infty\varphi(x)^2\,dx
=\frac2{\sqrt\pi}>1.
$$
The middle step is <integration by parts> with $\varphi'=-x\varphi$. Thus no proof of the displayed source inequality can hold for all its stated indices.
The correct <tail integral formula for expectation> is
$$
\mathbb E M_n=\int_0^\infty
\left[1-(2\Phi(x)-1)^{2^n}\right]\,dx.
$$
A valid <union bound> gives
$$
\mathbb P(M_n>x)\leq
\min\left(1,\frac{2^{n+1}}{\sqrt{2\pi}}\int_x^\infty e^{-u^2/2}\,du\right).
$$
In particular, for $x>1$ this is at most $\min(1,C_ne^{-x^2/2})$, where $C_n=2^{n+1}/\sqrt{2\pi}$. This corrected estimate proves the intended summability in part (c), without silently replacing the source formula.
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