What is the sum of this series '1+ e^y + e^2y = e^3y..... , ' where e^y <<1 . thanks!?

Question

really stuck on this- not even my maths teacher could help! Is it to do with the sum to infinity of an arithmetic or geometric series?

or actually just how would i find it out!

Answer 1

Assume, z=e^y

1+ e^y + e^2y + e^3y + ...
= 1+ z + z^2 + z^3 + ...
= (1 - z)^(-1) , since z<1
= 1/(1 - z)
= 1/(1 - e^y)

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Answer 2

It is a geometric progression with common ratio as e^y
Sum of infinite GP = a( 1-r^n) / 1 - r
here a is 1, r is e^y, n is infinity

apply:
Sum = 1. (1 - (e^y)^infinity ) / 1 - e^y
AS e^y is less than 1 so when the power is increased to infinity e^y becomes 0. ...(example 1/2 > (1/2)^2 > (12)^3 )
So sum = 1 (1-0) / 1 - e^y
= 1 / (1 - e^y)

Answer 3

A= 1/(1-e^y) e^y <1 if e^<<1 you can assume that the sum is 1