What is the sum from n=1 to infinity of 4(1/3)^{n-1} ?

The sum from n=1 to infinity of 4(1/3)^{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(1/3)^{n-1}

n = 1 \sum \infty 4 (\frac{1}{3})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 4 (\frac{1}{3})^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty (\frac{1}{3})^{n - 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

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

n = 1 \sum \infty [53 (\frac{1}{2})^{n}]

n = 0 \sum \infty \frac{8}{n ( n + 2 )}

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

n = 1 \sum \infty \frac{1}{5 ^{n} - 1}

What is the sum from n=1 to infinity of 4(1/3)^{n-1} ?
The sum from n=1 to infinity of 4(1/3)^{n-1} is converges