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

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

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

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

converges

Solution steps

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

Simplify

3^{n} \cdot \frac{1}{n !} : \frac{3 ^{n}}{n !}

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (e^{- 4 x} cos (4 x))

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y = 6 x^{2} - x^{3}

\frac{d y}{d x} = x + y - 1

\int (7 cos (x)) d x

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