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Researching on the internet, it is easy to find that Bernoulli was the first to give a one-digit approximation for $e$ (specifically, $2.5<e<3$). But, I cannot find any source describing exactly how he gave this approximation.

Some websites say he used a "power series" argument, but I cannot find this argument anywhere. A comment at the bottom of Bernoulli's Wikipedia article directs to Bernoulli's original paper (see the link below), but it is in Latin and the notation is not understandable to me.

https://books.google.com/books?id=s4pw4GyHTRcC&pg=PA222#v=onepage&q&f=false

Does anyone know of a source that has this original proof? Thank you for any help!

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  • $\begingroup$ Pretty sure that Bernoulli approximated $e$ by the limit: $$e=\lim_{n\to\infty} \left(1+\frac1n\right)^n$$ $\endgroup$ – Ali Caglayan Sep 7 '15 at 15:17
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    $\begingroup$ Which Bernoulli? $\endgroup$ – Bruno Joyal Sep 7 '15 at 15:18
  • $\begingroup$ Jacob Bernoulli. Yes, you are correct that Bernoulli used that limit, he was investigating compound interest. He showed that that limit is bounded by 2.5 and 3, and I am trying to understand his argument. $\endgroup$ – BlueBerry Sep 7 '15 at 15:24
  • $\begingroup$ I have seen in an article that the first person to approximate $e$ was a banker. $\endgroup$ – Oussama Boussif Sep 7 '15 at 15:37
  • $\begingroup$ @OussamaBoussif - Do you have a reference to that article? $\endgroup$ – uniquesolution Sep 7 '15 at 15:50
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Long comment

In the paper linked (1690 - see also : The number e ) Jacob Bernoulli is trying to solve the

problem of compound interest and, in examining continuous compound interest, he tried to find the limit of $(1 + 1/n)^n$ as $n$ tends to infinity. He used the binomial theorem to show that the limit had to lie between $2$ and $3$ so we could consider this to be the first approximation found to $e$.

In the latin text he is considering a sum $a$ with interest (usura annua) $b$, arriving at the following approximation :

summa major est quam $$a+b+ \frac{bb}{2a}$$

sed minor quam $$a+b+ \frac{bb}{2a-b}.$$

Then he put $a=b$ and the result is :

$$a+a+ \frac{a^2}{2a} < \text {summa} < a+a+ \frac{a^2}{2a-a}.$$

Si $a=b$, debitur plus quam $2 \frac 1 2 a$, & minus quam $3a$.

If we put $a=1$ we finally have :

$$1+1+\frac 1 2 < \text {summa} < 1+1+1.$$


But Bernoulii does not call it e :

Leonhard Euler introduced the letter $e$ as the base for natural logarithms, writing in a letter to Christian Goldbach of 25 November 1731. Euler started to use the letter $e$ for the constant in 1727 or 1728, in an unpublished paper on explosive forces in cannons, and the first appearance of $e$ in a publication was Euler's Mechanica (1736).

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  • $\begingroup$ Thank you, this helps! Do you know how to make sense of Bernoulli's original series, the one that looks like, $a+b+\frac{bb}{2a}+\frac{b_3}{2 \text{in} 3 aa}+\cdots$? I can't tell what this is the expansion for, no do I know what "2 in 3" means. $\endgroup$ – BlueBerry Sep 7 '15 at 18:44
  • $\begingroup$ @BlueBerry - see e note 7. $\endgroup$ – Mauro ALLEGRANZA Sep 7 '15 at 19:14
  • $\begingroup$ @BlueBerry - the exponent are written as "pedici"; i.e. it must be : $a +b + \frac {bb} {2a} + \frac {b^3} {2in3a^2} + \ldots$. $\endgroup$ – Mauro ALLEGRANZA Sep 7 '15 at 19:27
  • $\begingroup$ @BlueBerry - $2in3$ must be $2 \times 3$, because he speak of geometric progression ... $\endgroup$ – Mauro ALLEGRANZA Sep 7 '15 at 19:27
  • $\begingroup$ Thanks for the explanation. I'm still having difficulty seeing how this equation gives us the value of the account at the end of the year. Why is the original amount $a$ appearing in the denominator? $\endgroup$ – BlueBerry Sep 7 '15 at 19:49

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