= Solution
For $k\geq2$, write
$$
c_k=\frac1{k+1}-\frac1{2k}=\frac{k-1}{2k(k+1)},\qquad 0<c_k<\frac12.
$$
Use positivity from part (d) and bound the absolute terms of the <power series> in part (c) by a <geometric series>. For $n\geq2$,
$$
0<u_n\leq\sum_{k=2}^{\infty}\frac{c_k}{n^k}<\frac12\sum_{k=2}^{\infty}\frac1{n^k}=\frac1{2n^2(1-1/n)}\leq\frac1{n^2}.
$$
For $n=1$, $u_1=\tfrac32\ln2-1<\tfrac12<1$, since $\ln2=\int_1^2dt/t<1$. Thus a bound valid for every positive integer $n$ is
$$
\boxed{0<u_n<An^m,\qquad A=1,\quad m=-2.}
$$
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