= Solution
Put $M=\|f\|_\infty>0$ and $M_0=\mu(E)$. There is a positive-measure set on which $|f|>M/2$, so $M_0>0$ and each $M_n$ is positive; boundedness and finite measure make each finite. For $n\geq1$, apply <Cauchy-Schwarz> to $|f|^{(n-1)/2}$ and $|f|^{(n+1)/2}$ to get
$$
\boxed{M_n^2\leq M_{n-1}M_{n+1}.}
$$
Therefore the positive ratios $r_n=M_{n+1}/M_n$ are nondecreasing. Telescoping gives $M_n/M_0=r_0\cdots r_{n-1}\leq r_n^n$, while $|f|^{n+1}\leq M|f|^n$ almost everywhere gives $r_n\leq M$. Hence
$$
\boxed{\mu(E)^{-1/n}\|f\|_n\leq\frac{M_{n+1}}{M_n}\leq M.}
$$
For $0<\varepsilon<M$, the set $A_\varepsilon=\{|f|>M-\varepsilon\}$ has positive measure by the definition of <essential supremum>. Thus $\|f\|_n\geq(M-\varepsilon)\mu(A_\varepsilon)^{1/n}$, whereas $\|f\|_n\leq M\mu(E)^{1/n}$. Letting $n\to\infty$ and then $\varepsilon\downarrow0$ shows $\mu(E)^{-1/n}\|f\|_n\to M$. The preceding squeeze yields \b[$M_{n+1}/M_n\to\|f\|_\infty$]. This proves the ratio limit rather than assuming it from the moment-root limit.
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