I need to evaluate the upper and lower Riemann sums for $f(x) = 1/x$ on the interval $[1, 2]$ with division points $x_i = 2^{i/n}$ for $0 \leqslant i \leqslant n $. I need to verify that $\lim_{n \rightarrow \infty} U(f, P_n) = \ln 2 = \lim_{n \rightarrow \infty} L(f, P_n)$

All right. So I can write the lower sum as

$ \displaystyle L(f,P_n) = \sum^n_{i} \frac{1}{x_i - 1}\Delta x$, and the upper sum as $ \displaystyle U(f,P_n) = \sum^n_{i} \frac{1}{x_i}\Delta x$ if I'm not mistaken. But then to actually evaluate the sums, and later calculate the limits, I need to plug in the expression for $x_i$ and somehow end up with an expression that can be evaluated. It is here that I am stuck and any help is appreciated!


Note that the mesh spacing is not uniform. The length of the $i$'th subinterval is $x_i-x_{i-1}=2^{i/n}-2^{(i-1)/n}$.

Then, we can write

$$\begin{align} L(f,P_n)&=\sum_{i=1}^n \frac{1}{x_{i-1}}(x_i-x_{i-1})\\\\ &=\sum_{i=1}^n \left(\frac{1}{2^{(i-1)/n}}\right)\left(2^{i/n}-2^{(i-1)/n}\right)\\\\ &=\sum_{i=1}^n\left(2^{1/n}-1\right)\\\\ &=n\left(2^{1/n}-1\right)\\\\ &=n\left(e^{\frac1n\log(2)}-1\right)\\\\ &=\log(2)+O\left(\frac1n\right)\tag 1 \end{align}$$


$$\begin{align} U(f,P_n)&=\sum_{i=1}^n \frac{1}{x_{i}}(x_i-x_{i-1})\\\\ &=\sum_{i=1}^n \left(\frac{1}{2^{i/n}}\right)\left(2^{i/n}-2^{(i-1)/n}\right)\\\\ &=\sum_{i=1}^n\left(2^{1/n}-1\right)\\\\ &=n\left(1-2^{-1/n}\right)\\\\ &=n\left(1-e^{-\frac1n\log(2)}\right)\\\\ &=\log(2)+O\left(\frac1n\right)\tag 2 \end{align}$$

Taking the limit as $n\to \infty$ of $(1)$ and $(2)$, we find

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}L(f,P_n)=\lim_{n\to \infty}U(f,P_n)=\log(2)}$$

as was to be shown!

  • 1
    $\begingroup$ And feel free to also up vote as you see fit. $\endgroup$ – Mark Viola Apr 25 '17 at 4:45

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