Find an integer $N$ such that when $n≥N$, $ \left| U \left( \ln,P_n \right)-\int_1^2 {\ {\ln \left( x \right) }}\,dx \right|<ε\,\,? $ With 
$\,U \left( \ln,P_{{n}} \right) =\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \frac {1}{n}
$
and
$\,L \left( \ln,P_{{n}} \right) =\sum _{i=1}^{n}\ln  \left( 1+{\frac {i-1}{n}} \right) \frac {1}{n}
$
I get
$$
U \left( \ln,P_{{n}} \right)-L \left( \ln,P_{{n}} \right)={\frac {\ln  \left( 2 \right) }{n}}.
$$
For all $ε>0$, and $\,U \left( \ln,P_{{n}} \right) =\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \frac {1}{n}\,\,$, how do I proceed to show that there exists an integer $N$ such that when $n≥N$, when 
$$
\left| U \left( \ln,P_n \right)-\int_1^2 {\ {\ln  \left( x \right) }}\,dx \right|<ε\,\,?
$$
All help appreciated
 A: $\log(x)$ is a continuous and bounded function on the interval $(1,2)$, hence it is a Riemann integrable function and 
$$ \lim_{n\to +\infty}\left|U(f,P_n)-\int_{1}^{2}\log(x)\,dx\right|=0$$
follows from the very definition of Riemann integrability. Explicit bounds can be derived from the Hermite-Hadamard inequality, since $\log(x)$ is a concave function on $[1,2]$. In particular:
$$\frac{1}{n}\left[\frac{1}{2}\log(1)+\log\left(1+\frac{1}{n}\right)+\ldots+\log\left(1+\frac{n-1}{n}\right)+\frac{1}{2}\log(2)\right]\leq \int_{1}^{2}\log(x)\,dx $$
and
$$ \int_{1}^{2}\log(x)\,dx \leq \frac{1}{n}\left[\log\left(1+\frac{1}{n}\right)+\ldots+\log\left(1+\frac{n-1}{n}\right)+\log(2)\right]$$
is trivial, hence $U(f,P_n)-\int_{1}^{2}\log(x)\,dx$ is bounded by $\frac{\log(2)}{2n}$ in absolute value.
A: You were very close. Just recall that 
$$L(f,P_n) \le \int _1^2 \ln x\, dx \le U(f,P_n).$$
This says $\int _1^2 \ln x\, dx$ lies in the interval $[L(f,P_n),U(f,P_n)].$ Thus the distance from $\int _1^2 \ln x\, dx$ to the upper end point – and this is what you are estimating – is bounded above by the length of the interval, which, as you found, is $(\ln 2)/n.$ Therefore any $N$ such that $\frac{\ln 2}{N} <\epsilon$ will work.
