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

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

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

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

k \cdot 2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

2

= k \cdot 2

n = 2 \sum \infty 0.3 3^{n}

n = 3 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n - 1}

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

n = 0 \sum \infty \frac{6}{n}

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