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

The sum from k=1 to infinity of (k^4)/(k!) is converges

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sum from k=1 to infinity of (k^4)/(k!)

k = 1 \sum \infty \frac{k ^{4}}{k !}

converges

Solution steps

k = 1 \sum \infty \frac{k ^{4}}{k !}

Simplify

k^{4} : k^{3} k

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

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty n^{- 5}

n = 3 \sum \infty \frac{1}{n ^{2} - 4}

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

n = 1 \sum \infty (\frac{2}{13})^{n}

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