What is the sum from n=0 to infinity of 4/(3^{n-1)} ?

The sum from n=0 to infinity of 4/(3^{n-1)} 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 4/(3^{n-1)}

n = 0 \sum \infty \frac{4}{3 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{4}{3 ^{n - 1}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 0 \sum \infty \frac{1}{3 ^{n - 1}}

Apply Series Ratio Test:

converges

= 4 converges

= converges

x \to 0 lim (\frac{2}{e ^{x}})

\frac{d y}{d x} + 5 y = x e^{- 5 x}

\frac{8 x}{x ^{2} + 1}

(20, 8), (9, 16)

What is the sum from n=0 to infinity of 4/(3^{n-1)} ?
The sum from n=0 to infinity of 4/(3^{n-1)} is converges