What is the sum from n=0 to infinity of (9^n)/(n!) ?

The sum from n=0 to infinity of (9^n)/(n!) is converges

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sum from n=0 to infinity of (9^n)/(n!)

n = 0 \sum \infty \frac{9 ^{n}}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{9 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (3 x^{2} - 5)

y^{^{'}} = \frac{5 y}{x}

f (x) = 2 x^{2} + x, at x = 0

(1 + x^{2}) (e^{2 t} d t - e^{x} d x) - (1 + x) = 0

x \to π lim (\frac{x}{4} - \frac{π}{2})

What is the sum from n=0 to infinity of (9^n)/(n!) ?
The sum from n=0 to infinity of (9^n)/(n!) is converges