A Sigma Postulate

Sum of $\frac { 1 }{ { 2 }^{ n } } , n=1\quad to\quad \infty$

=> $\frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +\frac { 1 }{ { 2 }^{ 4 } } .................$

=> $\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } (1+.........$

Let $\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } (1+\frac { 1 }{ 2 } (1+......$ be $n$

$\therefore \frac { 1 }{ 2 } (1+n) = n$

$=>1+n = 2n$

$=>1=n$

sum of 1/2^n,n=1 to infinity is geometric progression(G.P.) with a=1/2 & r=1/2

formula for infinite summation=a/(1-r) when r<1

hence Sum=1