What is the sum from n=1 to infinity}((3^{n-1 of))/(9^n) ?

The sum from n=1 to infinity}((3^{n-1 of))/(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=1 to infinity}((3^{n-1 of))/(9^n)

n = 1 \sum \infty \frac{( 3 ^{n - 1} )}{9 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{3 ^{n - 1}}{9 ^{n}}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (4 x 16 - x^{2})

\int_{1}^{2} 2 x x^{2} + 1 d x

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

(\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 3 y^{3})

\int \frac{1}{e ^{u}} d u

What is the sum from n=1 to infinity}((3^{n-1 of))/(9^n) ?
The sum from n=1 to infinity}((3^{n-1 of))/(9^n) is converges