What is the sum from n=1 to infinity of n/(9^n) ?

The sum from n=1 to infinity of n/(9^n) is converges

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sum from n=1 to infinity of n/(9^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 4 \sum \infty \frac{3}{n ( n - 1 )}

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

i = 1 \sum \infty \frac{1}{3 ^{i}}

n = 3 \sum \infty \frac{3}{n ( n + 3 )}

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

What is the sum from n=1 to infinity of n/(9^n) ?
The sum from n=1 to infinity of n/(9^n) is converges