Write $S_k$ for a sum of $k$ of the iid terms; $E[S_k]=0$ , $\operatorname{Var}(S_k)=k\sigma^2$ . By Chebyshev, for any exponent $p$ ,

$P\Bigl(\Bigl|\frac{S_k}{n^p}\Bigr|>\varepsilon\Bigr) \le\frac{k\sigma^2}{\varepsilon^2n^{2p}},$

which $\to0$ whenever $p>1/2$ (since then $k/n^{2p}\lesssim n^{1-2p}\to0$ ). A single fixed term like $\exp(X_1)$ , divided by anything $\to\infty$ , also $\to0$ almost surely.

(A): exponent $1$ (on $n$ ) exceeds $1/2$ ; both the two fixed exponentials over $n$ and $S_{n-2}/n$ vanish. True.

(B): exponent $5/8=0.625>1/2$ , so by the Chebyshev bound above, $S_{n-1}/n^{5/8}\to0$ in probability, and $\exp(X_1)/n^{5/8}\to0$ too. So this statement is also true — it does converge to $0$ in probability, contrary to being "the false one."

(C) and (D) share the exponent $3/8=0.375<1/2$ : here $S_n/n^{3/8} =(S_n/\sqrt n)\cdot n^{1/2-3/8}=(S_n/\sqrt n)\cdot n^{1/8}$ . By the CLT, $S_n/\sqrt n$ converges in distribution to a non-degenerate $N(0,\sigma^2)$ , while $n^{1/8}\to\infty$ ; the product therefore grows without bound in probability (for any $M$ , $P(|S_n/n^{3/8}|>M)\to1$ ) — it neither settles to a constant (so (C)'s claim of non-convergence in probability is correct) nor approaches any proper limiting distribution (so (D)'s claim of convergence in distribution to a genuine random variable is false).

Our own computation therefore identifies (D) as the false statement, via a direct, standard Chebyshev bound for (B) and a routine CLT-scaling argument for (C)/(D); the officially published key marks (B) instead, which appears to be an error, since $5/8>1/2$ puts (B) squarely in the "vanishes" regime just like (A).