How to find Riemann Sum

Question

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

Answer 1

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.

Answer 2

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} $