Infinite sum of infinite negative series e^(-nt)

[thomas49th]

Find the sum of
e^(-nt) as n = 0,1,2,3,...

Apparently it's 1/(1-e^-t)

Can't use normal geometric series formula though as if a =1, r = e?
Any ideas
Thanks :)

 

[Curious3141]

Find the sum of
e^(-nt) as n = 0,1,2,3,...

Apparently it's 1/(1-e^-t)

Can't use normal geometric series formula though as if a =1, r = e?
Any ideas
Thanks :)

As long as t>0, you can treat this as the sum of an infinite geometric series.

The common ratio is e^(-t), which has magnitude less than 1. First term is e^0 = 1.

The sum is a/(1-r) = 1/(1-e^(-t)).

 

[thomas49th]

of course when the condition of a=1 occurs it's just 1/(1-r). it was very easy then. doh :)