Evaluate the geometric series or state that it diverges.

[infinity]
Σ e^-5n
n=0

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) )

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

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step-by-step explanation: