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I was curious what does $$\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{\vdots}}}}$$ evaluate to. Empirically, I observed that it equals approximately $0.5819767$, and a calculator found that this value agrees with $\frac{1}{e-1}$ to at least 8 places. Is $\frac{1}{e-1}$ the exact value of this continued fraction? If this is true, is this result new? And how could the equivalence be proved?

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    Great question, but it's a duplicate of This one. It does evaluate to $\frac{1}{e-1}$. Also see This question – Aniruddha Deb Jul 13 '20 at 13:35
  • I am not sure why your question is voted to be a duplicate of the one evaluating $\dfrac{1}{\text{e}-1}$. However, it is a duplicate of the one involving the lower gamma function. Observe that $$\gamma(1,1)=\int_0^1,\exp(-t),\text{d}t=1-\frac{1}{\text{e}},.$$ – Batominovski Jul 13 '20 at 14:11
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    Here is my old blog posting explaining how this continued fraction can be solved. – Sangchul Lee Jul 13 '20 at 14:21
  • @Batominovski Is it possible to prove this equality using elementary algebra (at least extreme elementary algebra, school level) ? –  Jul 13 '20 at 14:35
  • @Elementary I have absolutely no clue. I am not very knowledgeable about continued fractions. Sangchul Lee may be a better person for you to ask. – Batominovski Jul 13 '20 at 14:37
  • @SangchulLee Would you like to answer the question I asked Batominovski? –  Jul 13 '20 at 14:42
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    @Elementary, I am also not an expertise in continued fractions. Anyway, here is a solution based on the general theory: Let $$I_n=\frac{1}{(n-1)!}\int_{0}^{1}t^{n-1}e^t,\mathrm{d}t.$$ Then $I_n$ satisfies the following recurrence relation: $$I_1=e-1,\qquad I_2=1,\qquad I_n=nI_{n+1}+(n+1)I_{n+2}.$$ Indeed, it follows from integrating both sides of $(t^n(1-t)e^t)'=nt^{n-1}e^t-nt^ne^t-t^{n+1}e^t$ from $0$ to $1$. So we have $$\frac{1}{e-1}=\frac{1}{I_1/I_2}\qquad\text{and}\qquad I_n/I_{n+1}=n+\frac{n+1}{I_{n+1}/I_{n+2}}.$$ Repeatedly applying this relation then proves the identity. – Sangchul Lee Jul 13 '20 at 15:26
  • @SangchulLee Thank you for explanation. By the way Is this "logic" correct? The number $e$ is not algebraic. For this reason, the algebraic way doesn't exist for to prove this continued fraction.?Can we say? –  Jul 14 '20 at 22:09
  • I suspect that you have a particular technique in your mind, such as finding the value of a periodic continued fraction by setting up a rational equation and solving for it. Then yes, because $e$ is transcendental. – Sangchul Lee Jul 14 '20 at 22:20
  • @SangchulLee yes, That is what I mean...It is impossible to solve with a rational equation isn't it? this is what I want to ask.. like we've solved $\sqrt 2$ continued fraction. So looking for algebraic methods is meaningless, right? –  Jul 14 '20 at 22:28
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    I suspect so. At least we cannot obtain the value $\frac{1}{e-1}$ algebraically over $\mathbb{Q}$, although I am not sure if we can rule out the possibility in which your continued fraction may be associated algebraically with other transcendental quantities by some slick method. – Sangchul Lee Jul 14 '20 at 22:34
  • @SangchulLee Thank you. Your comment is point shot. –  Jul 14 '20 at 22:43

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