Based on previous, we can define tetra-logarithm as:
log[4](1+x)=1-x[4]2/2+x[4]3/3-x[4]5/4+....
and take x=(1/e)^2=0.13533528323661269189399949497248
then log[4](1+(1/e)^2)) = e^(1/e)-1=0.44466786101 (well I do not know for sure as series converge too slow for my methods).
Since log(1+(1/exp(1))^2)) = 0.126928011043
log[4] has smaller base than e.
To find it, e(4)^log[4](1+(1/e)^2))= e(4)^0.44466786101
0.44466786101 *ln(e(4)) = 0.126928011043
ln(e(4)) = 0.126928011043/0.44466786101 =0.285444535512
e(4)=exp^0.285444535512 = 1.33035328597
That should be the sum of 1+1/2^3+1/2^3^4 ..............?
Of course, it can not be true. I have driven to far into wrong direction, but idea was appealing.
Ivars
log[4](1+x)=1-x[4]2/2+x[4]3/3-x[4]5/4+....
and take x=(1/e)^2=0.13533528323661269189399949497248
then log[4](1+(1/e)^2)) = e^(1/e)-1=0.44466786101 (well I do not know for sure as series converge too slow for my methods).
Since log(1+(1/exp(1))^2)) = 0.126928011043
log[4] has smaller base than e.
To find it, e(4)^log[4](1+(1/e)^2))= e(4)^0.44466786101
0.44466786101 *ln(e(4)) = 0.126928011043
ln(e(4)) = 0.126928011043/0.44466786101 =0.285444535512
e(4)=exp^0.285444535512 = 1.33035328597
That should be the sum of 1+1/2^3+1/2^3^4 ..............?
Of course, it can not be true. I have driven to far into wrong direction, but idea was appealing.
Ivars