What is the sum from n=1 to infinity of (4^n)/(n!) ?

The sum from n=1 to infinity of (4^n)/(n!) is converges

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sum from n=1 to infinity of (4^n)/(n!)

n = 1 \sum \infty \frac{4 ^{n}}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{n !}

Apply Series Ratio Test:

converges

= converges

\int_{1}^{8} 3 \frac{ln ( x )}{x ^{2}} d x

(x^{2} + x y - y^{2}) d x + (x y) d y = 0

ln (x + 2), 3

x \to 0 lim (x \cdot ln (x^{2}))

\frac{d}{d t} (e^{t a n (t)})

What is the sum from n=1 to infinity of (4^n)/(n!) ?
The sum from n=1 to infinity of (4^n)/(n!) is converges