A good site documenting approximations of irrationals

Question

I'm thinking of Sloane here but I believe that only takes sequences/series into account.

Basically I've derived an interesting, appealing formula for e and want to know if it's already been discovered.

Answer 1

Wolfram Mathworld is a great site for information about irrational numbers. The article on e lists quite a few formulas along with other related topics.

Answer 2

Your formula appears to be related to the famous Bell numbers. So, even if not explicitly mentioned elsewhere, it is a rather trivial corollary to something quite well-known. See Dobinski's formula.

Answer 3

We want to show equality $$ e={2 \over 3} \left( {1 \over 1!} +{1 +2\over 2!} + {1 +2+3\over 3!} + {1 +2+3+4\over 4!} + ... \right) $$ Let's define a more general expression $f$ using the indeterminate $x$ at each summand inside the parenthese and derive the equality by finally inserting $x=1$: $$ \begin{array}{} f(x)&= {1 \over 1!} +{3\over 2!}x + {6\over 3!}x^2 + {10\over 4!}x^3 + ... \\ &={1 \over 2}\left( {2 \over 1!} +{6\over 2!}x + {12\over 3!}x^2 + {20\over 4!}x^3 + ... \right) \qquad \text{//cancel biggest factors}\\ &={1 \over 2}\left( {2 \over 0!} +{3\over 1!}x + {4\over 2!}x^2 + {5\over 3!}x^3 + ... \right) \qquad \text{//separate in two series} \\ &={1 \over 2}\left( {1 \over 0!} +{2\over 1!}x + {3\over 2!}x^2 + {4\over 3!}x^3 + ... \right) + {1 \over 2}\left( {1 \over 0!} +{1\over 1!}x + {1\over 2!}x^2 + {1\over 3!}x^3 + ... \right) \\ &={1 \over 2}\left( {1 \over 0!}x +{1\over 1!}x^2 + {1\over 2!}x^3 + {1\over 3!}x^4 + ... \right)'+ {1 \over 2}e^x \qquad \text{//note the first derivative!}\\ &={1 \over 2}\left( x e^x \right)'+ {1 \over 2}e^x\\ &={1 \over 2}\left( 1+ x \right)e^x+ {1 \over 2}e^x\\ &={1 \over 2}(2+x)e^x\\ \end{array} $$ and if we insert $1$ for $x$ in $f(x)$ we get equality $$ e = {2\over 3} f(1) ={2\over 3} \left( {1 \over 2} (2+1) \right)e^1=e $$