What is the sum from n=1 to infinity of (n^2)/(8^n) ?

The sum from n=1 to infinity of (n^2)/(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^2)/(8^n)

n = 1 \sum \infty \frac{n ^{2}}{8 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{8 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{9 ^{n}}{n + 1}

n = 0 \sum \infty (\frac{10}{11})^{n}

i = 0 \sum \infty \frac{1}{2 ^{i}}

n = 3 \sum \infty \frac{4}{7 ^{n}}

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

What is the sum from n=1 to infinity of (n^2)/(8^n) ?
The sum from n=1 to infinity of (n^2)/(8^n) is converges