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Using the Riemann Integral and the fact that the indefinite integral of $\frac{1}{x}$ is $\ln|x| + C$ I get

$$\int_a^b \frac{1}{x}dx = \lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{a+\Delta xi}\Delta x=(\ln|b|+C)-(\ln|a|+C)=\ln|b|-\ln|a|$$

where $\Delta x =\frac{b-a}{n}$

Taking $\Delta x$ out of the numerator and denominator

$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\Delta x} + i}\frac{\Delta x}{\Delta x}=\ln|b|-\ln|a|$$

$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\Delta x} + i}=\ln|b|-\ln|a|$$

Replacing $\Delta x$ with $\frac{b-a}{n}$

$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{a}{\frac{b-a}{n}} + i}=\ln|b|-\ln|a|$$

$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{an}{b-a} + i}=\ln|b|-\ln|a|$$

Substituting $a=1$

$$\lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{\frac{n}{b-1} + i}=\ln|b|-\ln|1|=\ln|b|$$

Finally, replacing $b=x$

$$\ln|x| = \lim_{n \to \infty} \sum_{i=1} ^{n} \frac{1}{i + \frac{n}{x-1}}$$

As far as I'm aware there is no known series of $\ln|x|$ where none of the terms have exponents. Could this particular series be of any use in the approximation of natural logarithm values, or any other application of the natural logarithm?

Jordan Tejada
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  • Please elaborate on your objection to integer powers and polynomial evaluation. – Lutz Lehmann Feb 23 '23 at 05:55
  • Possibly related to https://math.stackexchange.com/questions/46378/do-these-series-converge-to-logarithms – Jaume Oliver Lafont Feb 23 '23 at 05:57
  • It is Riemann, not Reimann! – Georg Friedrich Bernhard Riemann – Martin R Feb 23 '23 at 05:58
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    It should be noted that your result is in fact not a series because each term depends on $n$. Some partial answer to your question: I don't think that this representation is useful when calculating logarithms for two reasons:
    1. The series seems to converge rather slowly.
    2. If you want to calculate the sum up to $m > n$, you have to start from the beginning and cannot for example use the sum up to $n$ which you might have calculated before.
     – Bruno Krams Feb 23 '23 at 06:48

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