A better vaule for Napier's Constant? A Better value... ?

[2G7GE55U4]

Can you refute this statement?

An approximation of 'ℇ' is given by:
∞
'℮'~∑1/n!=1 + 1/1 + 1/1∙2 + 1/1∙2∙3 + 1/1∙2∙3∙4 + . . .
ⁿ⁼⁰
= 2.7182818284590452353602874713527...

But, a far better value for ℇ is given by
'ℇ' = ²⁷√ ( 6912275131084 / 13 )

= 2.7182184624259287300938728390698. . . !

because 1/'ℇ' elegantly solves the Hat Probability problem with:

1/'ℇ' = 0.36788801703139411761123754766767...!

 

[lookagain]

Can you refute this statement?

'℮'~∑1/n!=1 + 1/1 + 1/1∙2 + 1/1∙2∙3 + 1/1∙2∙3∙4 + . . .
ⁿ⁼⁰
= 2.7182818284590452353602874713527...

e is not approximately equal to, but it is equal to
the summation from n = 0 to oo of 1/n! =
1 + 1/1 + 1/(1*2) + 1/(1*2*3) + 1/(1*2*3*4) + ...

You were missing required grouping symbols around those denominators, and you showed
an infinite series, which in this case, sums to
e exactly.

 

[Otis]

Can you refute this statement? An approximation of 'ℇ' [is]
2.71828... [and a] far better value ...
2.71821...

I know the first 15 digits of e (from a silly mnemonic), but why is the second approximation above better than the first? Its fifth decimal digit is incorrect.

 

[topsquark]

I agree with Otis. Why is your [imath]\mathcal{E}[/imath] a better number than e, which is what the series is actually equal to? You can't be claiming that [imath]\mathcal{E}[/imath] is the same as [imath]\sum_{n = 0}^{\infty} \dfrac{1}{n!}[/imath] are you?

What does [imath]\mathcal{E}[/imath] do?

-Dan