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  convergence            \(\sum _{n=0}^{\infty }\frac{1}{2^n}\)

 

 

HELP PLS

 Feb 21, 2019

Best Answer 

 #1
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\(\sum \limits_{n=0}^\infty ~\alpha^n = \dfrac{1}{1-\alpha}\\ \text{here }\alpha = \dfrac 1 2\\ \text{and thus the sum is equal to }\dfrac{1}{1-\dfrac 1 2}= 2\)

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 Feb 21, 2019
 #1
avatar+4449 
+4
Best Answer

\(\sum \limits_{n=0}^\infty ~\alpha^n = \dfrac{1}{1-\alpha}\\ \text{here }\alpha = \dfrac 1 2\\ \text{and thus the sum is equal to }\dfrac{1}{1-\dfrac 1 2}= 2\)

Rom Feb 21, 2019

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