What is the sum from n=1 to infinity of (n^2)/(7^n) ?

The sum from n=1 to infinity of (n^2)/(7^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)/(7^n)

n = 1 \sum \infty \frac{n ^{2}}{7 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{7 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{4}{2 n + 3}

n = 0 \sum \infty 9^{- n}

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

n = 0 \sum \infty \frac{i}{n !}

n = 1 \sum \infty 5^{n} i

What is the sum from n=1 to infinity of (n^2)/(7^n) ?
The sum from n=1 to infinity of (n^2)/(7^n) is converges