What is the sum from n=1 to infinity of 4(0.6)^{n-1} ?

The sum from n=1 to infinity of 4(0.6)^{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 4(0.6)^{n-1}

n = 1 \sum \infty 4 (0.6)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 4 \cdot 0. 6^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 0. 6^{n - 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

e^{x} - x^{4}

\int \frac{1}{( 36 + x ^{2} ) ^{\frac{3}{2}}} d x

x \to 0 lim (\frac{tan ( 7 x )}{x})

\frac{\partial}{\partial x} (x^{x})

g (x) = 1 + 2 x

What is the sum from n=1 to infinity of 4(0.6)^{n-1} ?
The sum from n=1 to infinity of 4(0.6)^{n-1} is converges