What is the sum from n=0 to infinity of 8(1/2)^{n-1} ?

The sum from n=0 to infinity of 8(1/2)^{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 8(1/2)^{n-1}

n = 0 \sum \infty 8 (\frac{1}{2})^{n - 1}

converges

Solution steps

n = 0 \sum \infty 8 (\frac{1}{2})^{n - 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty (\frac{1}{2})^{n - 1}

Apply Series Ratio Test:

converges

= 8 converges

= converges

x^{2} y

\int 15 sin^{2} (θ) d θ

x \to 3 lim (\frac{3}{( x - 3 ) ^{2}})

\frac{d}{d x} (ln (\frac{x ^{2}}{5} + \frac{9}{x}))

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

What is the sum from n=0 to infinity of 8(1/2)^{n-1} ?
The sum from n=0 to infinity of 8(1/2)^{n-1} is converges