Harvard-MIT Mathematics Tournament

Solution:

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,
$$f(n+1) - f(n) = \log\left(\frac{n+1}{n}\right) - c\left(\frac{1}{n} - \frac{1}{n+1}\right) - d\left(\frac{1}{n^{2}} - \frac{1}{(n+1)^{2}}\right) - k\left(\frac{1}{n^{3}} - \frac{1}{(n+1)^{3}}\right) + O\left(\frac{1}{n^{4}}\right)$$
For the left hand side, $f(n+1) - f(n) = \frac{1}{n+1}$. By substituting $x = \frac{1}{n}$, the formula above becomes
$$\frac{x}{x+1} = \log(1+x) - c x^{2} \cdot \frac{1}{x+1} - d x^{3} \cdot \frac{x+2}{(x+1)^{2}} - k x^{4} \cdot \frac{x^{2} + 3x + 3}{(x+1)^{3}} + O\left(x^{4}\right)$$
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 $k x^{4} \cdot \frac{x^{2} + 3x + 3}{(x+1)^{3}}$ 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 $d = -\frac{1}{12}$.