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

The sum from n=0 to infinity of ((3^n))/2 is diverges

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 ((3^n))/2

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

diverges

Solution steps

n = 0 \sum \infty \frac{3 ^{n}}{2}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 0 \sum \infty 3^{n}

Apply Series Geometric Test:

diverges

= \frac{1}{2} diverges

= diverges

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

\frac{1}{x}

\frac{d x}{d t} = 6 - \frac{3 x}{500}

\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

\frac{\partial}{\partial x} (x^{\frac{2}{3}})

What is the sum from n=0 to infinity of ((3^n))/2 ?
The sum from n=0 to infinity of ((3^n))/2 is diverges