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### Proof of the Infinite Series That Calculates 'e'

```Date: 02/04/2004 at 15:56:36
From: Tarik Selbes
Subject: proof of the infinite series which gives the constant e

Is there a proof about this infinite series that gives the value of e:
1 + 1/1! + 1/2! + 1/3! + 1/4! + . . . + 1/n! where n goes to infinity?

```

```
Date: 02/04/2004 at 21:18:13
From: Doctor Rob
Subject: Re: proof of the infinite series which gives the constant e

Thanks for writing to Ask Dr. Math, Tarik!

There are two proofs that I know of.

1.  The function e^x has derivative equal to itself.  Then the
Maclaurin series for any function which can be differentiated as many
times as you like is

f(x) = f(0)/0! + f'(0)*x/1! + f"(0)*x^2/2! + f"'(0)*x^3/3! + ...

For f(x) = e^x, you have

e^x = f(x) = f'(x) = f"(x) = f"'(x) = ...
1 = f(0) = f'(0) = f"(0) = f"'(0) = ...

and the Maclaurin series for e^x is then

e^x = 1/0! + x/1! + x^2/2! + x^3/3! + x^4/4! + ...

Now set x = 1, and you get the series about which you asked.

2.  The definition of e is

e =  lim (1+1/n)^n
n->oo

Consider the binomial expansion for n = 1, 2, 3, 4, 5, ...

n
(1+1/n)^n = SUM C(n,i)*(1/n)^i
i=0

For i = 0, 1, 2, 3, ... one has

C(n,i)*(1/n)^i = n!/[(n-i)!*i!*n^i]
= (1)*(1-1/n)*(1-2/n)*...*(1-[i-1]/n)/i!

whose limit as n grows without bound is 1/i!.  Then

n
lim (1+1/n)^n =  lim  SUM C(n,i)*(1/n)^i
n->oo            n->oo i=0

oo
= SUM  lim  C(n,i)*(1/n)^i
i=0 n->oo

oo
e = SUM 1/i!
i=0

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
High School Sequences, Series
High School Transcendental Numbers

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