What is the sum from n=1 to infinity of (n^8)/(8^n) ?

The sum from n=1 to infinity of (n^8)/(8^n) is converges

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^8)/(8^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{e ^{n}}{n ^{6}}

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

i = 1 \sum \infty 2 (\frac{4}{i})^{3} + 9

n = 1 \sum \infty \frac{1}{n} (\frac{4}{5})^{n}

n = 1 \sum \infty cos (\frac{9}{n})

What is the sum from n=1 to infinity of (n^8)/(8^n) ?
The sum from n=1 to infinity of (n^8)/(8^n) is converges