How to find Riemann Sum Let $f \left( x \right) =\ln  \left( x \right)$ at $[1, 2]$ and let $P_{{n}}=${$1,{\frac {n+1}{n}}, {\frac {n+2}{n}},...,{\frac {2\,n-1}{n}}, 2$} where $  1\leq n $ is an integer. Find $U \left( f,P_{{n}} \right)$ and $L \left( f,P_{{n}} \right)$ and show that $U \left( f,P_{{n}} \right)-L \left( f,P_{{n}} \right)={\frac {\ln  \left( 2 \right) }{n}}$. 
I find
$$U \left( f,P_{{n}} \right) =\sum _{i=1}^{n}f \left( x_{{i}} \right) \Delta\,x_{{i}}=\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \frac {1}{n}$$
and
$$L \left( f,P_{{n}} \right) =\sum _{i=1}^{n}f \left( x_{{i-1}} \right) \Delta\,x_{{i}}=\sum _{i=1}^{n}\ln  \left( 1+{\frac {i-1}{n}} \right) \frac {1}{n}.$$
On the other hand is
$$U \left( f,P_{{n}} \right) ={\frac {1}{n}\ln  \left( {\frac {n+1}{n}} \right) }+{\frac {1}{n}\ln  \left( {\frac {n+2}{n}} \right) }+{\frac {1}{n}\ln  \left( {\frac {n+3}{n}} \right) }+...+{\frac {\ln  \left( 2 \right) }{n}}$$
and 
$$L \left( f,P_{{n}} \right) ={\frac {1}{n}\ln  \left( 1 \right) }+{\frac {1}{n}\ln  \left( {\frac {n+1}{n}} \right) }+{\frac {1}{n}\ln  \left( {\frac {n+2}{n}} \right) }+...+{\frac {1}{n}\ln  \left( {\frac {2n-1}{n}} \right) } $$
which gives the desired result
$$U \left( f,P_{{n}} \right)-L \left( f,P_{{n}} \right)={\frac {\ln  \left( 2 \right) }{n}}.$$
But still, I have not found 
$$U \left( f,P_{{n}} \right) =\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \frac {1}{n}$$
or
$$L \left( f,P_{{n}} \right) =\sum _{i=1}^{n}\ln  \left( 1+{\frac {i-1}{n}} \right) \frac {1}{n}.$$
All help appreciated
 A: As zipirovich wrote,
you should have
$\frac1{n}$,
not $\frac{i}{n}$.
To get the difference
is surprisingly easy.
$\begin{array}\\
U \left( f,P_{{n}} \right)
&=\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \frac {1}{n}\\
&=\frac1{n}\sum _{i=1}^{n}\ln  \left( 1+{\frac {i}{n}} \right) \\
\end{array}
$
and
$\begin{array}\\
L \left( f,P_{{n}} \right) 
&=\sum _{i=1}^{n}\ln  \left( 1+{\frac {i-1}{n}} \right) \frac {1}{n}\\
&=\frac1{n}\sum _{i=0}^{n-1}\ln  \left( 1+{\frac {i}{n}} \right) \\
&=\frac1{n}\left(\sum _{i=1}^{n}(\ln   1+{\frac {i}{n}})-\ln(2) \right) \\
&=U \left( f,P_{{n}} \right)-\frac{\ln(2)}{n} \\
\end{array}
$
A: First of all, both your "sigma" expressions for $U(f,P_n)$ and $L(f,P_n)$ have the same error in them: each $\displaystyle \Delta x_i=\frac{1}{n}$, not $\displaystyle \frac{i}{n}$.
Other than that, I'd say that you actually solved the problem. I think that setting up these summation expressions qualifies as finding $U(f,P_n)$ and $L(f,P_n)$. I doubt that they can be simplified further in any useful way.
