Let $f(x)=\ln(x)$ on $[1,2]$. Let the partitions be given as $P_n=(a=1,\frac{n+1}{n},\frac{n+2}{n},...,\frac{2n-1}{n},2=b)$ for $n\in \mathbb{Z} \geq1$
1) I want to calculate the upper and lower sums. I have $\Delta x=\frac1n$ and I'm using this formula (let me know if it's wrong):
Upper sum $U(f,P_n)$: $$f(c_i)=\ln(1+\frac{i}{n})$$ Taking the sum $$\frac{1}{n}\sum_{i=1}^n\ln(1+\frac{i}{n})$$
Lower sum $L(f,P_n)$ $$f(c_i)=\ln(1+\frac {i-1}{n})$$ $$\frac{1}{n}\sum_{i=1}^n\ln(1+\frac{i-1}{n})$$ I have tried calculating this for like two days. I keep getting the wrong result. (BTW I have been trying to confirm it by inserting for n=3 and comparing it with a Riemann sum calculator).
My problem is when I take the sum, I know that the $i$ becomes $n!$ but how can I calculate when I have different terms ($1+\frac{i-1}{n}$) in $\ln$? I could maybe write it as
$$\frac{1}{n}[\sum_{i=1}^n\ln(n+i-1)-\sum_{i=1}^n\ln(n+i)]$$ ?
2) Show that $$U(f,P_n)-L(f,P_n)=\frac{\ln(2)}{n}$$
Since $f$ is monotone I could use $$\Delta x \times[f(b)-f(a)]=\frac1 n[\ln 2 - \ln 1]=\frac{\ln(2)}{n}$$ Does that make sense? I haven't really used the upper and lower sums, so it seems "too easy".