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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y^{^{'}} + 3 y = 2 e^{- 3 x}

y = x^{3} - x, (- 1, 0)

f (x) = \frac{2}{x}, at x = - 1

\frac{1}{x} d y - \frac{y}{x ^{2}} d x = 0

n = 1 \sum \infty (\frac{2}{7})^{n}

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