Find the value of
1 + 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + …
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The partial sum with top term 1 + N 1 comes out to be ∑ k = 0 N k ! 1 after a bit of simplification. In the example displayed, with N = 5 , the sum is ∑ k = 0 5 k ! 1 = 6 0 1 6 3 . The limit is 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 is the "super number" and 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..........!!!!
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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 as a series of positive unit fractions, x = a 1 + a b 1 + a b c 1 + . . . , where each denominator is a multiple of the previous one. The number e is (or can be) defined by its Engel series, of course, e = ∑ k = 0 ∞ k ! 1 , the simplest Engel series there is.
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 , which can be expanded out into an infinite fraction.
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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.........!!!
I published a couple of problem s that are inspired by this. I bet they also might be related to that function.
This is simply the Engel expansion of Euler's number.......!!!
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Let us define a n = n 1 + n + 1 1 + n + 2 ⋯ and the value of the expression be x . Then
x = 1 + a 1 = 1 + 1 + a 2 = 1 + 1 + 2 1 + a 3 = 1 + 1 + 2 1 + 2 3 1 + a 4 = 1 + 1 + 2 1 + 3 ! 1 + 3 ! 4 1 + a 5 = 0 ! 1 + 1 ! 1 + 2 ! 1 + 3 ! 1 + 4 ! 1 + ⋯ = e ≈ 2 . 7 1 8