What is the sum from n=0 to infinity of i/(2^n) ?

The sum from n=0 to infinity of i/(2^n) is i*2

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sum from n=0 to infinity of i/(2^n)

n = 0 \sum \infty \frac{i}{2 ^{n}}

i \cdot 2

Solution steps

i = 0 \sum \infty \frac{n}{2 ^{i}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= i \cdot n = 0 \sum \infty \frac{1}{2 ^{n}}

Apply Series Geometric Test:

2

= i \cdot 2

n = 1 \sum \infty 2^{n + 1}

n = 0 \sum \infty n \cdot 2^{n}

n = 1 \sum \infty (- π)^{n}

n = 2 \sum \infty n e^{- 4 n}

n = 1 \sum \infty \frac{3}{n} - \frac{3}{n ^{2}}

What is the sum from n=0 to infinity of i/(2^n) ?
The sum from n=0 to infinity of i/(2^n) is i*2