What is the sum from n=1 to infinity of (n^4)/(2^n) ?

The sum from n=1 to infinity of (n^4)/(2^n) is converges

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sum from n=1 to infinity of (n^4)/(2^n)

n = 1 \sum \infty \frac{n ^{4}}{2 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{4}}{2 ^{n}}

Apply exponent rule:

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

\frac{n ^{4}}{2 ^{n}} = (n^{4}) 2^{- n}

= n = 1 \sum \infty n^{4} \cdot 2^{- n}

Apply Series Ratio Test:

converges

= converges

\int (3 - 2 x - x^{4}) d x

\int 10 sin (x) 2 x d x

\frac{d}{d x} (arcsec (x) + arccsc (x))

\frac{x}{x + 3}

y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 2

What is the sum from n=1 to infinity of (n^4)/(2^n) ?
The sum from n=1 to infinity of (n^4)/(2^n) is converges