What is the limit as n approaches infinity of (3^{n+1})/((n)!) ?

The limit as n approaches infinity of (3^{n+1})/((n)!) is 0

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limit as n approaches infinity of (3^{n+1})/((n)!)

n \to \infty lim (\frac{3 ^{n + 1}}{( n ) !})

0

Solution steps

n \to \infty lim (\frac{3 ^{n + 1}}{n !})

Simplify

3^{n + 1} : 3^{n} \cdot 3

= n \to \infty lim (\frac{3 ^{n} \cdot 3}{n !})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 3 \cdot n \to \infty lim (\frac{3 ^{n}}{n !})

Use the series convergence definition

: 0

= 3 \cdot 0

Simplify

= 0

y^{^{'}} = \frac{x ^{3} - 2 y}{x}

\int t^{2} + e^{t} d t

\int x^{3} e^{4 x} d x

\int_{0}^{π} 9 sin^{2} (4 x) d x

x \to 0 lim (sin^{3} (2 x))

What is the limit as n approaches infinity of (3^{n+1})/((n)!) ?
The limit as n approaches infinity of (3^{n+1})/((n)!) is 0