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

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

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

converges

Solution steps

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

Simplify

(\frac{4}{3})^{1 - n} : \frac{4}{3} (\frac{4}{3})^{- n}

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= \frac{4}{3} converges

= converges

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

n = 1 \sum \infty 10 (- \frac{7}{8})^{n}

n = 0 \sum \infty 4 n

n = 0 \sum \infty 12

n = 0 \sum \infty 20

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