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Calculate the sum of the geometric series \(1+\left(\frac{1}{5}\right)+\left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^3 + \dots\). Express your answer as a common fraction.

 Jan 5, 2021
 #1
avatar+50 
+3

To begin, note that this is working out as a sort of a pattern, where \((\frac{1}{5})^0 = 1, (\frac{1}{5})^1 = \frac15\), and then it continues on to where the exponential power keeps increasing by 1. Let's express this in sigma notation:

 

\(\sum _ {n=0} ^ {\infty} (\frac15)^n\), because there is no ending to the original sum, so we assume that it goes on forever to infinity. Now, we evaluate the sum, using the formula for geometeric series: \(a(\frac{1-r^n}{1-r})\), where we know a is the first term, which is 1. R is the common ratio, which we see that we are raising 1/5 to some power, so the common ratio is 1/5. Then, we have n, which is the number of terms. This is infinity, because the index of the summation ends at infinity. So, writing this out gives:

 

\(a(\frac{1-r^n}{1-r}) \implies 1(\frac{1-(\frac15)^{\infty}}{1-\frac15})\).

When raising something to the infinite power, we conclude that it is 0, so for the numerator, we have 1 - 0, which is 1, and for the denominator, we have 1 - 1/5, which is 4/5. 1 divided by 4/5 is 1 times 5/4, so the answer is \(\boxed{\frac54}\).

Please correct me if I made a mistake, I am always willing to learn! laugh

 Jan 5, 2021
 #2
avatar+303 
0

Good Job you are correct

 Jan 5, 2021
edited by hihihi  Jan 5, 2021
 #3
avatar+117576 
+1

THX, mobro....here's a slightly easier way

 

coomon ratio   = 1/5          a  = first term   = 1

 

Sum    =    a / (1 - r)  =  1  /( 1 -1/5)  =  1 / (4/5)  = 5/4

 

 

cool cool cool

 Jan 5, 2021

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