What is the sum from n=1 to infinity of n(1/4)^n ?

The sum from n=1 to infinity of n(1/4)^n is converges

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sum from n=1 to infinity of n(1/4)^n

n = 1 \sum \infty n (\frac{1}{4})^{n}

converges

Solution steps

n = 1 \sum \infty n (\frac{1}{4})^{n}

Apply Series Root Test:

converges

= converges

k = 1 \sum \infty \frac{5}{k + 6}

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 5}

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

n = 0 \sum \infty 20 (\frac{1}{4})^{n}

k = 2 \sum \infty (\frac{1}{3})^{2 k}

What is the sum from n=1 to infinity of n(1/4)^n ?
The sum from n=1 to infinity of n(1/4)^n is converges