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

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

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

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

n = 1 \sum \infty \frac{sin ( 1 )}{n}

n = 1 \sum \infty \frac{n}{6 n ^{3} + 5}

n = 0 \sum \infty 5 n^{2}

n = 2 \sum \infty \frac{1}{3 n + 4}

n = 1 \sum \infty 3 (0.4)^{n - 1}

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