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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?

otah007
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    http://math.stackexchange.com/questions/1670508/solve-lim-n-to-infty-left-frac1n1-frac1n2-cdots-frac1nn?rq=1 here you can find a nice argument to conclude from the line $\lim_{n \to \infty}\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ – Jose Paternina Sep 25 '16 at 23:09
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    @JosePaternina I don't think there is any way to conclude without $\log'(x) = 1/x$ – reuns Sep 25 '16 at 23:11
  • Wait, $\ln 2$ instead of $\infty$? I think you meant to use $f(x)=\frac{1}{1+x}$. Also, your manipulation of the limit is completely wrong because your sum-to-product ratio is a sum of reciprocals of products of all but one factor, not a sum of reciprocals of factors. – J.G. Mar 09 '19 at 10:06
  • @J.G. I'm 'integrating' from 1 to 2, so I don't want $f(x)=\frac{1}{1+x}$. And yes, you are correct that my sum/product quotient is wrong. – otah007 Mar 10 '19 at 16:45
  • @JosePaternina Thanks, that gives quite a few different solutions, as well as the one below. When I first posted this question three years ago, I was an A-Level student trying to understand where calculus came from. Now I'm an undergrad having just done a term of real analysis, so I can actually understand the Darboux integral in the link you posted. Makes me realise how far I've come :) – otah007 Mar 10 '19 at 16:48

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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

Elias Assaf
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