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Question Number 2791 by Filup last updated on 27/Nov/15

Knowing that e = \underset{i = 1}{\sum^{\infty}} \frac{1}{i!},

Show that e is finite .

That is, show the following is true :

S = {\exists x \in R :∣ x ∣< \infty, e = x}

Where S is the solution

Commented by prakash jain last updated on 27/Nov/15

Euler number e = \underset{i = 0}{\sum^{\infty}} \frac{1}{i!}

Answered by prakash jain last updated on 27/Nov/15

e = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . .

e = 1 + \dots \Rightarrow e > 1

remain terms

\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots

2! = 2

3! > 2^{2} \Rightarrow \frac{1}{3!} < \frac{1}{2^{2}}

Similary \frac{1}{4!} < \frac{1}{2^{3}}

\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots < \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} = \frac{1 / 2}{1 - 1 / 2} = 1

e < 1 + 1 = 2

1 < e < 2

Hence S is true .

Note that e in your question is not THE

constant e .

Commented by Filup last updated on 27/Nov/15

O h . Its a typo . It was meant to be the

constant . Nice proof, r e g a r d l e s s!

Commented by prakash jain last updated on 27/Nov/15

For the constant e, 2 < e < 3