Prove $\int^{n+1}_1\frac{1}{x} dx \leq 1 + 1/2 + \cdots + 1/n$. 
Source: https://math.mit.edu/~choiks/Pset%202%20solutions.pdf


I don't understand the third line $\int^{n+1}_1\frac{1}{x}dx  \leq 1 + 1/2 + \cdots + 1/n$.
I think it should rather be $\int^{n+1}_1\frac{1}{x}dx  \geq 1 + 1/2 + \cdots + 1/n$ because
$\int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$ and $1 + 1/2 + \cdots + 1/n$ is just addition of $f(x) = 1/x$ at certain values (integers). Therefore:
$1 + 1/2 + \cdots + 1/n < \int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$.
Thank you!

There are simple diagrams that compare sums and integrals.... Here what I call $f(x)$ is real valued and continuous, also positive. Things change (order of the inequalities) depending on $f$ increasing or decreasing. I guess $a < b$ are fixed integers.
if we have $f(x) > 0$ and $f'(x) > 0,$  then
$$ \int_{a-1}^{b} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a}^{b+1} \;  f(x) \; dx   $$
if we have $f(x) > 0$ but $f'(x) < 0,$  then
$$ \int_a^{b+1} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a-1}^b \;  f(x) \; dx   $$
 We use this second one, as we are going to integrate $1/x$ we take $a=2$ along with $b=n$
$$ \int_2^{n+1} \; \frac{1}{x} \;  dx \;  < \; \sum_{j=2}^n \;  \frac{1}{j} \; < \; \int_{1}^n \;  \frac{1}{x} \; dx   $$
$$ 1+ \int_2^{n+1} \; \frac{1}{x} \;  dx \;  < \; 1+ \sum_{j=2}^n \;  \frac{1}{j} \; < \; 1+ \int_{1}^n \;  \frac{1}{x} \; dx   $$
$$ 1+ \log (n+1) - \log 2 \;  < \;  \sum_{j=1}^n \;  \frac{1}{j} \; < \; 1+ \log n    $$
