Slog(Exponential Factorial(x))

[tommy1729]

ok I have a proof.

it does indeed converge and really fast !

slog( EF(x) ) = sum ( slog( EF(x) ) - slog( EF(x-1) ) ) + Constant. 

difference operator and sum operator cancel.
( telescoping if you want )

slog( EF(x) ) = x + y

now consider how going from slog( EF(x) ) to slog ( EF(x+1) ) changes.


EF(x+1) = (x+1)^EF(x)

so 

slog EF(x+1) = slog( (x+1)^x^(x-1)^... ) ) = slog( (x+1)^EF(x) )

slog( (x+1)^x^(x-1)^... ) ) = slog( (x+1)^EF(x) ) = slog( exp( ln(x+1) * x^(x-1)^... ) ) 

= 1 + slog( ln(x+1) * EF(x) )

= 2 + slog( ln(ln(x+1)) + ln(EF(x) ) )

= 3 + slog ( ln ( ln(ln(x+1)) + ln(EF(x) ) ) )

we know for large L and small S ( small compared to L ) :
 
log ( L + S ) < log ( L ) + 2 S/L.

now clearly slog ( a + b ) < slog(a) + slog(b)

so

3 + slog ( ln ( ln(ln(x+1) + ln(EF(x) ) ) < 3 + slog( ln^[2](EF(x)) + o(1/sqrt (EF(x)) ) )

applying slog ( a + b ) < slog(a) + slog(b)

< 3 + slog(ln^[2](EF(x))) + slog( o(1/sqrt EF(x) ) )

< 1 + slog(EF(x)) + slog( o(1/sqrt EF(x) ) )

< 1 + x + y + slog( o(1/sqrt EF(x) ) )

So

1 < slog( EF(x+1) ) - slog( EF(x) ) < 1 + slog( o(1/sqrt EF(x) ) )

therefore 

slog( EF(x) ) = sum ( slog( EF(x) ) - slog( EF(x-1) ) ) + Constant < constant + sum ( 1 + slog( o(1/sqrt EF(x) ) ) )

< constant + x + constant2.

so slog( EF(x) ) - x converges to some constant. And fast.

a similar conclusion can be drawn from another base since slog(b,x) - slog(e,x) also converges to a constant.


regards

tommy1729