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

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

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sum from n=1 to infinity of n*(1/3)^n

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

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

n = 1 \sum \infty \frac{n ^{2}}{9 ^{n}}

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

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

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

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