What is the sum from n=1 to infinity of n(1/7)^n ?

The sum from n=1 to infinity of n(1/7)^n is converges

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sum from n=1 to infinity of n(1/7)^n

n = 1 \sum \infty n (\frac{1}{7})^{n}

converges

Solution steps

n = 1 \sum \infty n (\frac{1}{7})^{n}

Apply Series Root Test:

converges

= converges

n = 1 \sum \infty \frac{( 3 n - 2 )}{n}

n = 2 \sum \infty (\frac{2}{5})^{n}

n = 1 \sum \infty \frac{4 ^{n}}{n ^{7}}

k = 1 \sum \infty \frac{1}{k ^{7}}

n = 0 \sum \infty \frac{5 ^{n}}{7 ^{n}}

What is the sum from n=1 to infinity of n(1/7)^n ?
The sum from n=1 to infinity of n(1/7)^n is converges