What is the sum from n=1 to infinity of (n!)/(12^n) ?

The sum from n=1 to infinity of (n!)/(12^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=1 to infinity of (n!)/(12^n)

n = 1 \sum \infty \frac{n !}{1 2 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{n !}{1 2 ^{n}}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty \frac{n ^{2}}{7 ^{n}}

k = 1 \sum \infty (\frac{1}{4})^{k}

n = 1 \sum \infty \frac{8}{e ^{n} - 1}

n = 0 \sum \infty \frac{1}{2 n ^{3}}

n = 0 \sum \infty \frac{n}{( n + 1 ) ^{2}}

What is the sum from n=1 to infinity of (n!)/(12^n) ?
The sum from n=1 to infinity of (n!)/(12^n) is diverges