Exploring Pentation - Base e

[Ivars]

I took

I added asymtotic values of negative infinite of base e heptation and [9] to the sum I mentioned before:

Sum [5] =1/1,85035452902718^8-1/(2*1,8503545290271^7+1/(3*1,8503545290271^6-1/(4*1,8503545290271^5+1/(5*1,8503545290271^4=0,007297583=1/137,0316766

From Andrew's graph, I found the values to be roughly e[7]-infinity = -3,751 and e[9]-infinity = -5,693.

Then I put them in the same sum, obtaining:

Sum [7] =1/3,751^8-1/(2*3,751)^7+1/(3*3,751)^6-1/(4*3,751)^5+1/(5*3,751)^4=3,20285E-05

Sum[9] = 1/5,693^8-1/(2*5,693)^7+1/(3*5,693)^6-1/(4*5,693)^5+1/(5*5,693)^4=3,15992E-05

Then I made Sum [5,7,9] = Sum[5]-Sum[7]+Sum[9] = 0,007297583-3,20285E-05+3,15992E-05=0,0072971534=1/137,039738252

So after this, approximation of alpha =0.07297352570(5) got even better, as I expected, but of course I do not know the exact values of e[7]-infinity and e[9]-infinity and more.

Then we could see how does the sum Sum[5]-Sum[7]+Sum[9]-Sum[11]+Sum[13]-Sum[15]+.........converge.

Ivars