What is the sum from k=1 to infinity of (3k^8)/(k!) ?

The sum from k=1 to infinity of (3k^8)/(k!) is converges

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sum from k=1 to infinity of (3k^8)/(k!)

k = 1 \sum \infty \frac{3 k ^{8}}{k !}

converges

Solution steps

k = 1 \sum \infty \frac{3 k ^{8}}{k !}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot k = 1 \sum \infty \frac{k ^{8}}{k !}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\frac{\partial}{\partial x} (ln (x y^{4}))

y = 3 x e^{2 x}

\int - 1120 (1 + 4 t)^{- 3} d t

f (x) = 5 x^{2} - x^{3}

\int_{\frac{π}{2}}^{π} 3 x - sin (x) d x

What is the sum from k=1 to infinity of (3k^8)/(k!) ?
The sum from k=1 to infinity of (3k^8)/(k!) is converges