We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+2
77
1
avatar+194 

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

 

 

HELP PLS

 Feb 21, 2019

Best Answer 

 #1
avatar+5088 
+4

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

.
 Feb 21, 2019
 #1
avatar+5088 
+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

5 Online Users