I wish to calculate the integral $\int_1^2 \frac{1}{x}$ using Darboux sum using the division $p_n=2^{\frac{j}{n}}, 0\le j\le n$
I tried calculating the upper sum:
$$U(f,p_n)=\sum_{j=1}^n(2^{\frac jn}-2^{\frac{j-1}{n}})({2^{-\frac{j-1}{n}}})= \sum_{j=1}^n(2^{\frac 1n}-1)=n(2^{\frac{1}{n}}-1)$$
This should converge to $\ln(2)$ although I don't know how to show it.
have I done it right? how should I proceed?
continuing from where I stopped:
$U(f,p_n)= \sum_{j=1}^n(2^{\frac 1n}-1)=n(2^{\frac{1}{n}}-1 )$
Taking the limit I get: $\lim_{n\rightarrow \infty}n(2^{\frac{1}{n}}-1 )=\lim_{x\rightarrow 0}\frac{2^x-1}{x}=\ln(2)$
Using the same idea I get $L(f,p_n)$ is also $\ln(2)$
Therefore $U(f,p_n)=L(f,p_n)$ so the function is integratable and: $$\int_1^2 \frac{1}{x}=ln(2)$$