Continued Factorial Fractions

Calculus Level 1

1 + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \ldots What is the value of this continued fraction correct up to 2 decimal places?


This problem is a part of set nested radicals

1.61 1.53 3.14 2.71

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5 solutions

By definition, e e is given by the Taylor Expansion given in the question. Hence, answer is e e , approximately 2.71 2.71 .

Luis Valdivia
Mar 22, 2015

Applying the binomial theorem to the limit definition of e, one gets 1/0!+1/1!+1/2!+1/3!+1/4!+…

Therefore the answer is e.

Samarth Agarwal
Mar 31, 2015

The given series is the expansion of e x e^x . Here x=1. Thus, the value of series is e 1 e^1 \Rightarrow 2.71

Brock Brown
Mar 27, 2015

Python 2.7:

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from math import factorial as fac
from fractions import Fraction as frac
from time import time
total = 0
denominator = 0
end = time() + 10
while time() < end:
    total += frac(1,fac(denominator))
    denominator += 1
print "Answer:", float(total)

Aryan Gaikwad
Mar 21, 2015

The number is e e (Euler's Number). It is a irrational number, so you cannot define its exact value but you can get close to it.

You can either get a rough value by calculating the value of first few fractions (This works because the value of each fraction starts decreasing as you move towards right and after a few calculations, the value almost becomes negligible) -

1 + 1 + 1 2 + 1 6 + 1 24 + 1 120 = 2.718055556 1+1+\frac { 1 }{ 2 } +\frac { 1 }{ 6 } +\frac { 1 }{ 24 } +\frac { 1 }{ 120 } =2.718055556

Or you can calculate a more accurate value of it using this formula -

( 1 + 1 n ) n { (1+\frac { 1 }{ n } ) }^{ n }

where n n is a very large number. As n n approaches infinity, the expression above approaches e e .

Exactly.. e+2 XD

Mehul Arora - 6 years, 2 months ago

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