What is the sum from n=1 to infinity of n^4e^{-n} ?

The sum from n=1 to infinity of n^4e^{-n} is converges

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sum from n=1 to infinity of n^4e^{-n}

n = 1 \sum \infty n^{4} e^{- n}

converges

Solution steps

n = 1 \sum \infty n^{4} e^{- n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{\frac{7}{6}}}

n = 0 \sum \infty 10 + 2 n

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

n = 1 \sum \infty \frac{1}{7 ^{n - 1}}

n = 1 \sum \infty ln (1 + n)

What is the sum from n=1 to infinity of n^4e^{-n} ?
The sum from n=1 to infinity of n^4e^{-n} is converges