Question

Find the sum of the infinite series: 1 + 1/2 + 1/2² + 1/2³ + ...
     (Hint: Let Sn = 1 + 1/2 + 1/2² + 1/2³ +...+ 1/2^n
             Multiply, Sn by 2 to get the following equation:
            2Sn = 2 + 1 + 1/2 + 1/2² +...+ 1/2^n + 1/2^(n+1)
            Subtract Sn from 2Sn and let Sn tend to infinity to get the sum)

Answer 1

Sum of the infinite series: 1 + 1/2 + 1/2² + 1/2³ + ...

Using power series
Σ x^n / 2^n = (x/2) / [1 - (x/2)] = x/(2-x)

Then differentiating
Σ n x^(n-1)/ 2^n = 2/(2-x)^2

Plugging in x = 1
Σ n / 2^n = 2

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