= Solution
{c}
Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\geq2$. The <factorial> gives
$$
\frac{u_{n+1}}{u_n}=\frac{11}{4(n+1)}\frac{\log(n+1)}{\log n}.
$$
The <logarithm> identity
$$
\frac{\log(n+1)}{\log n}=1+\frac{\log(1+1/n)}{\log n}\longrightarrow1
$$
follows from <continuity> of the <logarithm> at one and $\log n\to+\infty$. The product rule for <limits of sequences> therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\leq u_n/2$ for all $n\geq N$. By <mathematical induction>, $0<u_{N+j}\leq u_N2^{-j}$. A <geometric sequence> with ratio in $(0,1)$ tends to zero, so the <squeeze theorem> proves
$$
\boxed{\lim_{n\to\infty}\frac{11^n\log n}{2^{2n+1}n!}=0.}
$$
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