An Unexpected Number

Calculus Level 3

Find the value of

1 + 1 + 1 + 1 + 1 + 1 + 5 4 3 2 1 \displaystyle 1+\cfrac{1 + \cfrac{1 + \cfrac{1 + \cfrac{1 + \cfrac{1+\dots}{5}}{4}}{3}}{2}}{1}


The answer is 2.7182818284.

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3 solutions

Chew-Seong Cheong
Nov 19, 2018

Let us define a n = 1 + 1 + n + 2 n + 1 n a_n = \dfrac {1+ \frac {1+\frac \cdots{n+2}}{n+1}}n and the value of the expression be x x . Then

x = 1 + a 1 = 1 + 1 + a 2 = 1 + 1 + 1 + a 3 2 = 1 + 1 + 1 2 + 1 + a 4 3 2 = 1 + 1 + 1 2 + 1 3 ! + 1 + a 5 4 3 ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + = e 2.718 \begin{aligned} x & = 1 + a_1 \\ & = 1 + 1 + a_2 \\ & = 1 + 1 + \frac {1+a_3}2 \\ & = 1 + 1 + \frac 12 + \frac {\frac {1+a_4}3}2 \\ & = 1 + 1 + \frac 12 + \frac 1{3!} + \frac {\frac {1+a_5}4}{3!} \\ & = \frac 1{0!} + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \frac 1{4!} + \cdots \\ & = e \approx \boxed{2.718} \end{aligned}

Otto Bretscher
Nov 18, 2018

The partial sum with top term 1 + 1 N 1+\frac{1}{N} comes out to be k = 0 N 1 k ! \sum_{k=0}^{N} \frac{1}{k!} after a bit of simplification. In the example displayed, with N = 5 N=5 , the sum is k = 0 5 1 k ! = 163 60 \sum_{k=0}^{5} \frac{1}{k!}=\frac{163}{60} . The limit is e \boxed{e} .

Very nice puzzle, but I would claim that the appearance of Euler's number is never totally "unexpected." ;) One of my profs used to say that e e is the "super number" and e x e^x is the "super function."

Yeah, Sir...........that is essentialy how I solved this, then searched for it on the net to find out its name...........And that is how I got to know about the Engel expansions..........!!!!

Aaghaz Mahajan - 2 years, 6 months ago

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Yes, the Engel series expansion is an interesting concept. Friedrich Engel was interested in the (almost) unique representation of a positive real number x x as a series of positive unit fractions, x = 1 a + 1 a b + 1 a b c + . . . x=\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}+... , where each denominator is a multiple of the previous one. The number e e is (or can be) defined by its Engel series, of course, e = k = 0 1 k ! e=\sum_{k=0}^{\infty} \frac{1}{k!} , the simplest Engel series there is.

Otto Bretscher - 2 years, 6 months ago

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What do you mean by "(almost) unique"?

Julian Poon - 2 years, 6 months ago

Interesting, I didn't know there was a name to this. I came up with this problem trying to solve the functional equation f ( x + 1 ) = x f ( x ) 1 f(x+1) = xf(x) - 1 , which can be expanded out into an infinite fraction.

Julian Poon - 2 years, 6 months ago

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Yeah!!! There are a lot of things which I come to know about after searching them........even if I sometimes think that they have no names.........!!!

Aaghaz Mahajan - 2 years, 6 months ago

I published a couple of problem s that are inspired by this. I bet they also might be related to that function.

Blan Morrison - 2 years, 6 months ago

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@Blan Morrison Ok thanks......!!! :)

Aaghaz Mahajan - 2 years, 6 months ago
Aaghaz Mahajan
Nov 18, 2018

This is simply the Engel expansion of Euler's number.......!!!

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