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

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SUMMARY

The infinite sum of the series e^(-nt) for n = 0, 1, 2, 3,... converges to 1/(1 - e^(-t)) when t > 0. This series can be treated as an infinite geometric series where the first term a is 1 and the common ratio r is e^(-t), which satisfies the condition |r| < 1. The standard formula for the sum of a geometric series, S = a/(1 - r), applies here, confirming the result.

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thomas49th
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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 :)
 
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thomas49th said:
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)).
 
of course when the condition of a=1 occurs it's just 1/(1-r). it was very easy then. doh :)
 

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