What is the sum from n=1 to infinity of (n^2)/(3^n) ?

The sum from n=1 to infinity of (n^2)/(3^n) is converges

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sum from n=1 to infinity of (n^2)/(3^n)

n = 1 \sum \infty \frac{n ^{2}}{3 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{3 ^{n}}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (x^{n})

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f (x) = 3 tan (3 x), at x = \frac{π}{18}

\int 3 sin^{5} (x) cos (x) d x

What is the sum from n=1 to infinity of (n^2)/(3^n) ?
The sum from n=1 to infinity of (n^2)/(3^n) is converges