What is the sum from n=1 to infinity of (n^2)/(9^n) ?

The sum from n=1 to infinity of (n^2)/(9^n) is converges

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sum from n=1 to infinity of (n^2)/(9^n)

n = 1 \sum \infty \frac{n ^{2}}{9 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{9 ^{n}}

Apply Series Ratio Test:

converges

= converges

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

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

n = 1 \sum \infty \frac{1}{n ^{2}} + \frac{1}{n}

n = 0 \sum \infty \frac{6}{1 - n ^{2}}

n = 1 \sum \infty \frac{n - 1}{n 2 ^{n}}

What is the sum from n=1 to infinity of (n^2)/(9^n) ?
The sum from n=1 to infinity of (n^2)/(9^n) is converges