I showed previously that $f(x)=\ln x$ on $[1,2]$ has the upper sum $$U(f,P_n)=\frac 1 n\sum_{i=1}^{n}\ln\left(1+\frac in\right)$$
I want to prove that $\forall\varepsilon>0$, $\exists N\in\mathbb{Z}$ such that when $n\geq N$ we have $$\left|U(f,P_n)-\int_1^2\ln x dx\right|<\varepsilon$$ So I see I need to show that the distance between the upper sum and the integral is small from a certain point. I know we have that as $n$ the number of partitions increases $U$ gets smaller. So that the upper sum converges to the true value of the integral (and thus also closer to the lower sum). It makes sense than that the distance between the upper sum and the integral would become small from a certain point. I also know of the result that (let the definite integral be $I$ and the Riemann sum be $S(f,P_n))$:
$$\left|S(f,P_n)-I \right|<\varepsilon$$ Which also adds to what I'm trying to prove.
Moving along. I calculated $\int_1^2\ln x dx$ to be $2\ln 2-1$) $$\left|\frac 1 n\sum_{i=1}^{n}\ln\left(1+\frac in\right)-2\ln 2-1\right|<\varepsilon$$
Kinda stuck here
Note that $$ \int_1^2\ln xdx=\int_0^1\ln(1+x)dx=\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\ln(1+x)dx, U(f,P_n)=\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\ln(1+\frac{i}{n})dx$$ and that $f(x)=\ln(1+x)$ is increasing. So by the Lagrange MVT, one has \begin{eqnarray} &&|U(f,P_n)-\int_1^2\ln xdx|\\ &=&\bigg|\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\bigg[\ln(1+\frac{i}{n})-\ln(1+x)\bigg]dx\bigg|\\ &=&\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\bigg[\ln(1+\frac{i}{n})-\ln(1+x)\bigg]dx\\ &=&\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\frac{1}{1+t_i}(\frac{i}{n}-x)dx\\ &\le&\sum_{i=1}^n\int_{\frac{i-1}{n}}^{\frac{i}{n}}\frac{1}{1+t_i}(\frac{i}{n}-\frac{i-1}{n})dx\\ &\le&\sum_{i=1}^n\frac{1}{n}\int_{\frac{i-1}{n}}^{\frac{i}{n}}1dx\\ &=&\frac{1}{n}. \end{eqnarray} Thus for $\forall \epsilon>0$, there is $N\in\mathbb{N}$ such that when $n\ge N$, one has $\frac{1}{n}<\epsilon$. This implies $$ |U(f,P_n)-\int_1^2\ln xdx|<\epsilon. $$