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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (ln (x^{2}))

\frac{\partial}{\partial y} (x^{4} y + x z - y z^{2} - 7)

\frac{d y}{d x} = e^{6 y} e^{6 x}

\frac{d y}{d x} = y (y^{5} x + 2)

1 - 4 x^{2} y^{^{'}} = x

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