What is the sum from k=1 to infinity of (9k^3)/(k!) ?

The sum from k=1 to infinity of (9k^3)/(k!) is converges

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sum from k=1 to infinity of (9k^3)/(k!)

k = 1 \sum \infty \frac{9 k ^{3}}{k !}

converges

Solution steps

k = 1 \sum \infty \frac{9 k ^{3}}{k !}

Apply the constant multiplication rule:

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

= 9 \cdot k = 1 \sum \infty \frac{k ^{3}}{k !}

Apply Series Ratio Test:

converges

= 9 converges

= converges

n = 0 \sum \infty \frac{1}{3 n ^{2} + 8}

n = 0 \sum \infty n \cdot 0. 5^{n}

n = 1 \sum \infty \frac{1}{∣ n ∣}

n = 0 \sum \infty (\frac{- 1}{7})^{n}

n = 1 \sum \infty 8 (0.3)^{n - 1}

What is the sum from k=1 to infinity of (9k^3)/(k!) ?
The sum from k=1 to infinity of (9k^3)/(k!) is converges