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!

Induction does the trick
Assume it true for $n-1$, i.e.
$$1 + \ldots + \frac{1}{n-1} - \ln n < 1 - \frac{1}{n}$$
Then
$$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < \ln n + 1 - \frac{1}{n} + \frac{1}{n} - \ln (n+1) = 1 + \ln (\frac{n}{n+1}) = 1 + \ln(1 - \frac{1}{n+1})$$
Note that $$\ln (1-x) < -x$$ for $0 < x < 1$, hence 
$$ 1 + \ln(1 - \frac{1}{n+1}) < 1 - \frac{1}{n+1}$$ 
We conclude 
$$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < 1 - \frac{1}{n+1}$$
