Let $s_{n} := 1 + 1/2 + \cdots + 1/n - \ln(n+1)$ and show that $s_{n} \leq 1 - 1/(n+1)$. This is in a Riemann integration section of a practice sheet from real analysis. I know that if $f(x) := \frac{1}{[x]} - \frac{1}{x+1}$, where $[*]$ is the floor function, then $\int_{1}^{n} f = s_{n}$. But I'm not sure where to go to get the quantity on the right-hand side of the inequality. I feel that I should compare $f \leq g$ for some $g$ where $\int_{1}^{n} g = 1- \frac{1}{n+1}$, but I am having trouble doing that!

We have
$$s_n=\sum_{k=1}^{n}\frac1k-\ln(n+1)=\sum_{k=1}^{n}\left(\frac1k-\int_k^{k+1}\frac{dt}{t}\right)\le \sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=1-\frac1{1+n}$$
where the last equality is obtained by telescoping.
