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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

diverges

= 3 diverges

= diverges

\int (\frac{1}{x} - \frac{6}{x ^{2} + 1}) d x

\int_{0}^{16} x - \frac{1}{4} x d x

\frac{d}{d x} (\frac{4 x + 5}{8})

y = \frac{3}{x}, (5, \frac{3}{5})

\frac{s - 1}{s ( s + 1 ) ( s + 2 )}

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