So a few days ago I just kind of messed around with my calculator, when I had an idea about a new continued fraction. I inputted it, and I found that it converged really quickly, and, quite wondrously, converged to the base of the natural logarithm. Then, the day later, I tried computing a similar continued fraction, this time starting with the number 2 instead of 1, without the increment of the numerator by 1.

My first computation My second computation

Interestingly, the second computation converges to EXACTLY the same values, except it's faster in the sense that the first method is delayed by 1 value per time. Does anybody possibly have an intuition / explanation as for why this might be the case? I've looked at the general ways of computing e: the sum of integer factorials and (1+1/n)^n, which I suspect might be more relevant to this.

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    $\begingroup$ I feel like the first one can be rewritten as $\sum_{i=0}^\infty \frac{1}{i!}$ somehow. Then, the second sum is $2+\sum_{i=2}^\infty \frac{1}{i!}$, which is equivalent to the first sum. $\endgroup$ – Noble Mushtak Dec 29 '18 at 17:09
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    $\begingroup$ Both are just forms of the factorial series. (Why images, not formulas?) $\endgroup$ – metamorphy Dec 29 '18 at 17:10
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    $\begingroup$ The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series. $\endgroup$ – timtfj Jan 29 '19 at 20:20

Your second computation is equal to your first one: $$1 + \frac{2+\frac{3+\frac{4+\frac{5+ \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} = 1 + \frac{2}{2} + \frac{\frac{3}{3} + \frac{\frac{4}{4} + \frac{\frac{5}{5} + \frac{\frac{6}{6} + \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} = 2 + \frac{1+\frac{1+\frac{1+\frac{1+ \frac{\cdots}{\cdots}}{5}}{4}}{3}}{2} $$

I hope this is clear.


Your first calculation can be written as


which expands to

$$1+\frac{2}{2}+\frac{3}{2\cdot 3}+\frac{4}{2\cdot 3 \cdot 4} + \dots$$

Cancelling the repeated number in each term turns it into

$$1+\frac11+\frac1{1\cdot2}+\frac{1}{1\cdot2\cdot 3} + \dots$$

$$=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots$$ $$=e$$

You've effectively factorised the series and written the result as a big fraction.

  • $\begingroup$ Thank you for the explanation! Yes, I find this answer more useful than most other answers here. $\endgroup$ – Avtarás Karîm Elymés̱er Jan 30 '19 at 21:45

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