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