= Solution
Part (a) gives $\mathbb P(X\in\mathbb Z)=0$. Therefore the <ceiling function> and <floor function> satisfy $\lceil X\rceil=\lfloor X\rfloor+1$ almost surely. Both are integrable, since $0\le\lfloor X\rfloor\le X$ and $\lceil X\rceil\le X+1$. Thus \b[$\mathbb E[\lceil X\rceil]=\zeta(t-1)+1$].
Now take expectations in $\lfloor X\rfloor\le X\le\lceil X\rceil$:
$$
\zeta(t-1)\le\frac{t-1}{t-2}\le\zeta(t-1)+1.
$$
Writing $s=t-1$, which ranges over every $s>1$, and rearranging yields
$$
\boxed{\frac1{s-1}\le\zeta(s)\le\frac{s}{s-1}\qquad(s>1).}
$$
These are the <zeta bounds from rounded Pareto variables>: rounding below and above converts one elementary <expected value> into bounds on an infinite <series>.
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