Answer:
sum = = 1 / ( 1-e^(-5) )
( 1.00678 to 5 decimals)
Step-by-step explanation:
Question: sum
[infinity]
Σ e^-5n
n=0
We note that this is a geometric series,
with n=0, e^(-5n) = e^0 = 1
with n=1, e^(-5n) = e^(-5)
with n=2, e^(-5n) = e^(-10)
...
Therefore the common ratio is r=e^(-5)
and the sum for a geometric series
a+ar+ar^2 + ... ar^N
=a(1-r^(N+1)) / (1-r)
=1(1-r^(infinity-1)) / (1-e^(-5))
= 1 / ( 1-e^(-5) )
