What is the sum from n=1 to infinity of n/(3^{n-1)} ?

The sum from n=1 to infinity of n/(3^{n-1)} is converges

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sum from n=1 to infinity of n/(3^{n-1)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty - 8 (- \frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{3 ^{n + 1}}

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

x = 0 \sum \infty \frac{1}{5 ^{x}}

n = 1 \sum \infty \frac{1}{1. 1 ^{n}}

What is the sum from n=1 to infinity of n/(3^{n-1)} ?
The sum from n=1 to infinity of n/(3^{n-1)} is converges