Generalised Formulae for e

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of

e = {1 + (1 + k)/1! + (1 + 2k)/2! + (1 + 3k)/3! + (1+ 4k)/4! + …}/(1 + k),

where k = 0.

So, when k = 1,

e = {1 + 2/1! + 3/2! + 4/3! + 5/4! + …}/2

And when for example k = 10,

e = {1 + 11/1! + 21/2! + 31/3! + 41/4! + …}/11

This relationship holds for all real values of k (including non-integers) except – 1.

 

The case of k = – 1 is fascinating as it suggests that,

e ={(1 + 0/1! – 1/2! – 2/3! – 3/4! – …)}/0

And the limiting value of the expression inside the brackets as the number of terms

n ⁓ ∞ = 0.

So in this case the suggested limiting value of 0/0 = e?

 

Then,

e^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

So when x = 1,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …

 

And when x = 2,

e² = 2⁰/0! + 2¹/1! + 2²/2! + 2³/3! + 2⁴/4! + …

 

Therefore, combining the two formulae,

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + …,

is a special case of,

e^x(1 + 2k) = x⁰/0! + x¹(1 + k)/1! + x²(1 + 2k)/2! + x³(1 + 3k)/3! + x⁴(1 + 4k)/4! + …

for all x where x = 1 and k = 0.

Then when k = 0,

e ^x = x⁰/0! + x¹/1! + x²/2! + x³/3! + x⁴/4! + …

And when also x = 1,

e = 1⁰ + 1¹/1! + 1²/2! + 1³/3! + 1⁴/4! + …

However, if for example k = 2 and x = 2,

Then 5e² = 1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …

e ² = {1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5

so that,

e = [{1⁰/0! + (3*2¹)/1! + (5*2²)/2! + (7*2³)/3! + (9*2⁴)//4! + …}/5]^1/2