Preface: I'm a A-Level student, so much of the maths I'm speaking about here is quite new to me, in particular Riemann Sums. I apologise if this already has an answer, I couldn't find it.
I'm trying to calculate the integral of 1/x from first principles using a Riemann Sum. Using this formula:
$$\lim_{n \to \infty}\sum_{i=1}^{n}f\left(x_i\right)\Delta{x}=\int_{0}^{1}f\left(x\right)dx$$
I tried to find the integral between x=1 and x=2 like this:
$$f\left(x\right)=\frac{1}{x},a=1,b=2\therefore\Delta{x}=\frac{b-a}{n}=\frac{1}{n},f\left(x_i\right)=\left(a+i\Delta{x}\right)^{-1}=\frac{1}{1+\frac{i}{n}}$$
$$\therefore\lim_{n \to \infty}\sum_{i=1}^{n}f\left(x_i\right)\Delta{x}=\lim_{n \to \infty}\sum_{i=1}^{n}\frac{1}{n+i}$$
$$=\lim_{n \to \infty}\frac{1}{n+1}+\frac{1}{n+2}+…+\frac{1}{n+n}$$
$$=\lim_{n \to \infty}\frac{\sum_{i=1}^n(n+i)}{\Pi_{i=1}^n(n+i)}$$
$$=\lim_{n \to \infty}\frac{n^2+\frac{n\left(n+1\right)}{2}}{\frac{\left(2n\right)!}{n!}}$$
$$=\lim_{n \to \infty}\frac{\frac{3n^2+n}{2}}{\frac{\left(2n\right)!}{n!}}$$
$$=\lim_{n \to \infty}\frac{n!n\left(3n+1\right)}{2\left(2n\right)!}$$
And that's as far as I can go. I know the answer is $\ln2$, and testing what I've got so far in an Excel spreadsheet shows that it does indeed converge to $\ln2$ as $n$ gets larger.
Where do I go from here?
Best Answer
you can use $Xj = 2^{j/n}$ and $\Delta Xj = (2^{j/n}-2^{(j-1)/n)}$. you get $$\sum_{j=1}^{n} f(xj)*\Delta xj$$ which is $$\sum_{j=1}^{n} 2^{-j/n} * (2^{j/n}-2^{(j-1)/n})$$
and you do some extra math and you get $$\sum_{j=1}^{n} (1-2^{-1/n}) = n*(1-2^{-1/n}) = \frac{2^{-1/n}-1}{-1/n}$$ which when you take the limit n to infinity, 1/n goes to 0, that gets you ln(2), https://www.emathzone.com/tutorials/calculus/limit-of-ax-1x.html