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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

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

n = 0 \sum \infty 5 (- \frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{3 + e ^{- n}}

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

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

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