A reciprocal area and the endpoint choice

The width is $1/n$. Right endpoint $x_r=1+r/n$ has height $1/(1+r/n)=n/(n+r)$. Multiplying by the width gives rectangle area $1/(n+r)$, so $S_n=\sum_{r=1}^n1/(n+r)$.

The rectangles converge to the area $\int_1^2\frac1x\,dx=[\ln|x|]_1^2=\ln2-\ln1=\ln2$.

At each $x=1+r/n$ for $r=0,\ldots,n-1$, draw a horizontal top to the next partition point and vertical sides down to the axis. The first height is $f(1)=1$ and first area is $1/n$, as the Teacher notes. The left total is $L_n=\sum_{r=0}^{n-1}\frac1{n+r}$. Since $1/x$ decreases, the left rectangles overestimate and the right rectangles underestimate: $S_n<\ln2<L_n$. Both limits are $\ln2$.