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I read this QA which proves formula (I didn't dig into Euler's proof which uses 3 assumed theorems for point 40 to 42): $$ e-2 = 0+\cfrac1{1+\cfrac1{2 + p}} = 0+\cfrac1{1+\cfrac1{2 + \cfrac2{3 + \cfrac3{4 + \cfrac4{5 + \cfrac5{6 + \cfrac6{7 + \cfrac7{\cdots}}}}}}}} $$

OEIS referred to in mathworld gives: $$ e-2 = 0+\cfrac1{1+\cfrac1{2+\cfrac1{1+\cfrac1{1+\cfrac1{4+\ldots}}}}} $$

The paper referred to in OEIS proves using convergent. I can understand the proof assuming we know the recurrence relation for convergents which I didn't dig into its proof by the paper reference 4.

I tried to prove the 2nd is equal to the 1st by simplifying $p$ using division to $$ \cfrac1{1+k}=\cfrac1{1+\cfrac1{2} + \cfrac{\cfrac3{2}}{4 + \cfrac4{5 + \cfrac5{6 + \cfrac6{7 + \cfrac7{\cdots}}}}}} $$ can't be easily simplified due to $k$ can't be easily simplified.

Is there one elegant to prove these 2 continued fractions are equal without using $e-2$ as the medium?

An5Drama
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    Most non-simple fractions are usually not easy to transform in the simple fraction form, so my guess is no. – jjagmath Jun 26 '24 at 12:05
  • @jjagmath Thanks. I am learning SICP now where section 1.3.3 Exercise 1.38 refers to the above 2nd formula. I will return to this question if I read one book about continued fractions possibly. – An5Drama Jun 27 '24 at 05:27

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