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

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

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

n = 1 \sum \infty \frac{9 ^{n}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{9 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

x \to \infty lim (3 x + x^{2})

(0, 3), (3, 0)

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2}})

\int x \cdot e^{x} d x

9 - y^{2}

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