What is the sum from n=0 to infinity of (n!)/(6^n) ?

The sum from n=0 to infinity of (n!)/(6^n) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (n!)/(6^n)

n = 0 \sum \infty \frac{n !}{6 ^{n}}

diverges

Solution steps

n = 0 \sum \infty \frac{n !}{6 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

x \to 1 lim (- x - 8)

\frac{d}{d x} ((e^{x}) sin (x))

(\frac{1}{x - 1})^{^{'}}

t cos (3 t) + 2 e^{- 4 t}

\int e^{- 3 x} sin (5 x) d x

What is the sum from n=0 to infinity of (n!)/(6^n) ?
The sum from n=0 to infinity of (n!)/(6^n) is diverges