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

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

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

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

converges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{n ^{2}}{3 ^{n}} = (n^{2}) 3^{- n}

= n = 2 \sum \infty n^{2} \cdot 3^{- n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty 3 (0.3)^{n - 1}

n = 1 \sum \infty \frac{5 n ^{3}}{6 ^{n}}

k = 0 \sum \infty (\frac{1}{3})^{k}

n = 0 \sum \infty \frac{2}{3 - n}

n = 1 \sum \infty - 3 (\frac{8}{7})^{n - 1}

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