What is the sum from n=1 to infinity of 2(0.8)^{n-1} ?

The sum from n=1 to infinity of 2(0.8)^{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=1 to infinity of 2(0.8)^{n-1}

n = 1 \sum \infty 2 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 2 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

ln (x^{3})

\frac{d}{d x} (a x^{3} + b x^{2} - \frac{c}{3} x + d)

y = x^{2} (x - 1)

\frac{\partial}{\partial x} (x^{7} y^{9} - x^{6} y^{8})

\int \frac{1}{( x ^{2} + x + 1 )} d x

What is the sum from n=1 to infinity of 2(0.8)^{n-1} ?
The sum from n=1 to infinity of 2(0.8)^{n-1} is converges