What is the sum from k=1 to infinity of 2/3 (1.5)^{k-1} ?

The sum from k=1 to infinity of 2/3 (1.5)^{k-1} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=1 to infinity of 2/3 (1.5)^{k-1}

k = 1 \sum \infty \frac{2}{3} (1.5)^{k - 1}

diverges

Solution steps

k = 1 \sum \infty \frac{2}{3} \cdot 1. 5^{k - 1}

Apply the constant multiplication rule:

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

= \frac{2}{3} \cdot k = 1 \sum \infty 1. 5^{k - 1}

Apply Series Ratio Test:

diverges

= \frac{2}{3} diverges

= diverges

f (x) = cos (3 x^{2})

\frac{x}{x + 7}

y^{^{'}} = \frac{3 x ^{2} 16 + y ^{2}}{y}

\frac{d}{d t} (t e^{3 t})

\frac{d}{d x} (e^{- \frac{3}{x}})

What is the sum from k=1 to infinity of 2/3 (1.5)^{k-1} ?
The sum from k=1 to infinity of 2/3 (1.5)^{k-1} is diverges