The point of the formula is to give something correct to more places than you can compute. To understand it rewrite 9^(-(4^(67))) as (3^2)^(-(4^(67))) = 3^(-(24^(67))). This is the same as 3^(-(2^85)). So you have an expression of the form

(1+1/n)^n

where n=3^(-(2^85)). To estimate the error between this and E, take logs and see how close we get to 1, then use that to get the multiplicative factor by which we might be off. So the log is of the form n*Log[1+1/n] with n again as above. Expand as a series, truncate at second order, this is n*(1/n-1/(2*n^2)+...) or 1-1/(2*n)+error where error<<1/n. Exponentiating shows our result is E*Exp[-1/(2*n)+error], now use what we know about n to show how close this comes to E.

Thank you Mariusz and Daniel.

I do understand the derivation of the formula.

However, since the Mma documentation talks about “arbitrary precision”, I guess I thought that the system would, upon being given something as ill-behaved as this, emit a partial result to machine precision (i.e. 10^308) or just run for a “long time” and then emit an approximation to much less than 10^24 digits (the magnitude of the “near miss”).

Mma doumentation says 2*$MaxMachineNumber converts to “arbitrary precision”.

I guess this is actually (1+”too small number” raised to large power) raised to a “too large number”) because 9^-4^42 underflows.

N[E,10^5] result shown immediately

N[E,10^10] computation result $Aborted after 1+ hour

I guess I do not understand the implementation limits of “arbitrary precision” in Mma.

Ref:

http://www.johndcook.com/blog/2014/03/30/amazing-approximation-to-e/

http://mathworld.wolfram.com/eApproximations.html