= Solution
{c}
Use the valid <Gaussian maximum> estimate from part (b), with
$$
a_n=\sqrt{2(n+1)\log2}>1.
$$
Split the <tail integral formula for expectation> at $a_n$. Bounding the first part by $a_n$ and applying the Gaussian tail estimate to the second gives
$$
\mathbb E M_n
\leq a_n+C_n\int_{a_n}^\infty e^{-x^2/2}\,dx
\leq a_n+\frac{C_n}{a_n}e^{-a_n^2/2}
=a_n+\frac1{\sqrt{2\pi}\,a_n}.
$$
Thus $\mathbb E M_n=O(\sqrt n)$. The finitely many small indices also have finite expectation; the bound above already covers them.
By <Tonelli's theorem>, nonnegativity permits interchanging sum and expectation:
$$
\boxed{\mathbb E\left[\sum_{n\geq1}2^{-n/2}M_n\right]
=\sum_{n\geq1}2^{-n/2}\mathbb E M_n<\infty.}
$$
The final series converges because its exponential decay dominates the factor $\sqrt n$. This establishes the requested conclusion independently of the false bound printed in (1).
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