10. $[\mathbf{8}]$ Let $f(n)=\sum_{k=1}^{n} \frac{1}{k}$. Then there exists constants $\gamma, c$, and $d$ such that
$$
f(n)=\ln (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+O\left(\frac{1}{n^{3}}\right),
$$
where the $O\left(\frac{1}{n^{3}}\right)$ means terms of order $\frac{1}{n^{3}}$ or lower. Compute the ordered pair $(c, d)$.
Answer: $\left(\frac{1}{2},-\frac{1}{12}\right)$ From the given formula, we pull out the term $\frac{k}{n^{3}}$ from $O\left(\frac{1}{n^{4}}\right)$, making $f(n)=$ $\log (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+\frac{k}{n^{3}}+O\left(\frac{1}{n^{4}}\right)$. Therefore,
$$
f(n+1)-f(n)=\log \left(\frac{n+1}{n}\right)-c\left(\frac{1}{n}-\frac{1}{n+1}\right)-d\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)-k\left(\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}\right)+O\left(\frac{1}{n^{4}}\right) \text {. }
$$

For the left hand side, $f(n+1)-f(n)=\frac{1}{n+1}$. By substituting $x=\frac{1}{n}$, the formula above becomes
$$
\frac{x}{x+1}=\log (1+x)-c x^{2} \cdot \frac{1}{x+1}-d x^{3} \cdot \frac{x+2}{(x+1)^{2}}-k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}+O\left(x^{4}\right) .
$$

Because $x$ is on the order of $\frac{1}{n}, \frac{1}{(x+1)^{3}}$ is on the order of a constant. Therefore, all the terms in the expansion of $k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}$ are of order $x^{4}$ or higher, so we can collapse it into $O\left(x^{4}\right)$. Using the Taylor expansions, we get
$$
x\left(1-x+x^{2}\right)+O\left(x^{4}\right)=\left(x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\right)-c x^{2}(1-x)-d x^{3}(2)+O\left(x^{4}\right) .
$$

Coefficient comparison gives $c=\frac{1}{2}$ and $d=-\frac{1}{12}$.