And it's pretty much [the definition of Euler's number](https://www.investopedia.com/terms/e/eulers-constant.asp), when you calculate compounding interest over infinitely small intervals!

Isn't the product equal to $$e-1$$? Otherwise, good job, as usual.

south_korea_ln

science

[Daily puzzle] A short break (and some answers)

$$
1 + \frac{1}{1} + \frac{1}{1}\frac{1}{2} + \frac{1}{1}\frac{1}{2}\frac{1}{3} + ...
$$

$$
1 + \sum_{n=1}^{\infty}{   \prod_{m=1}^{n}{ \frac{1}{m}  } }
$$

1 + [$$e - 1$$](https://www.wolframalpha.com/input?i2d=true&i=Sum%5B%5C%2840%29Product%5BDivide%5B1%2Cm%5D%2C%7Bm%2C1%2Cn%7D%5D%5C%2841%29%2C%7Bn%2C1%2C%E2%88%9E%7D%5D) = $$e$$

