What is the sum from n=0 to infinity of n/(7^n) ?

The sum from n=0 to infinity of n/(7^n) is converges

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sum from n=0 to infinity of n/(7^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 5 \sum \infty \frac{1}{( n - 2 ) ^{4}}

n = 2 \sum \infty 9 e^{- 2 n}

n = 0 \sum \infty e^{2 l n (n)}

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

n = 0 \sum \infty 5 (\frac{3}{π})^{n}

What is the sum from n=0 to infinity of n/(7^n) ?
The sum from n=0 to infinity of n/(7^n) is converges