= Solution
{c}
By part <6/a/iii>, $X$ is an <integer> with probability zero. Thus $\lceil X\rceil=\lfloor X\rfloor+1$ almost surely, and
$$
\boxed{\mathbb E[\lceil X\rceil]=\zeta(t-1)+1.}
$$
The pointwise inequalities $\lfloor X\rfloor\leq X\leq\lceil X\rceil$ imply
$$
\zeta(t-1)\leq\frac{t-1}{t-2}\leq\zeta(t-1)+1.
$$
Putting $s=t-1>1$ and rearranging yields
$$
\boxed{\frac1{s-1}\leq\zeta(s)\leq\frac s{s-1}.}
$$
This extracts bounds for an infinite <series> from simple rounding inequalities.
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