What is the sum from n=0 to infinity of n/(9^n) ?

The sum from n=0 to infinity of n/(9^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=0 to infinity of n/(9^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{2 n}{5 n - 3}

x = 1 \sum \infty x e^{- x^{2}}

n = 3 \sum \infty \frac{n}{5 ^{n}}

k = 1 \sum \infty k^{\frac{1}{k}}

n = 1 \sum \infty \frac{1}{n ^{3} + 5}

What is the sum from n=0 to infinity of n/(9^n) ?
The sum from n=0 to infinity of n/(9^n) is converges