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})

\int_{0}^{2} x^{5} \cdot e^{x^{2}} d x

\int (sin^{4} (x)) (cos^{2} (x)) d x

t an g e n t f (x) = 3 tan (3 x), at x = \frac{π}{18}

d er i v a t i v e f (x) = (8 x (x^{2} + 6)^{3})

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