What is the sum from n=0 to infinity of 7(2/3)^{n+2} ?

The sum from n=0 to infinity of 7(2/3)^{n+2} 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 7(2/3)^{n+2}

n = 0 \sum \infty 7 (\frac{2}{3})^{n + 2}

converges

Solution steps

n = 0 \sum \infty 7 (\frac{2}{3})^{n + 2}

Apply the constant multiplication rule:

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

= 7 \cdot n = 0 \sum \infty (\frac{2}{3})^{n + 2}

Apply Series Ratio Test:

converges

= 7 converges

= converges

n = 1 \sum \infty \frac{e ^{\frac{8}{n}}}{n}

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

n = 0 \sum \infty e^{- 7 n}

n = 1 \sum \infty \frac{6}{n ^{2} - 1}

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

What is the sum from n=0 to infinity of 7(2/3)^{n+2} ?
The sum from n=0 to infinity of 7(2/3)^{n+2} is converges