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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty (\frac{2}{13})^{n}

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

n = 0 \sum \infty (4 + n^{2})^{- 1}

k = 0 \sum \infty 1^{k}

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

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