What is the sum from n=0 to infinity of i/(3^n) ?

The sum from n=0 to infinity of i/(3^n) is i 3/2

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sum from n=0 to infinity of i/(3^n)

n = 0 \sum \infty \frac{i}{3 ^{n}}

i \frac{3}{2}

Solution steps

i = 0 \sum \infty \frac{n}{3 ^{i}}

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{2}

= i \frac{3}{2}

n = 0 \sum \infty \frac{n}{1 - n}

k = 2 \sum \infty (\frac{4}{5})^{3 k}

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

n = 1 \sum \infty 12 (- \frac{5}{8})^{n}

n = 2 \sum \infty 7 (- \frac{3}{8})^{n}

What is the sum from n=0 to infinity of i/(3^n) ?
The sum from n=0 to infinity of i/(3^n) is i 3/2