What is the sum from n=1 to infinity of 6/(3n+2) ?

The sum from n=1 to infinity of 6/(3n+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=1 to infinity of 6/(3n+2)

n = 1 \sum \infty \frac{6}{3 n + 2}

diverges

Solution steps

n = 1 \sum \infty \frac{6}{3 n + 2}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{3 n + 2}

Apply Series Integral Test:

diverges

= 6 diverges

= diverges

n = 4 \sum \infty \frac{e ^{- n} + 1}{n}

n = 0 \sum \infty \frac{6}{n ( n + 2 )}

k = 1 \sum \infty \frac{k}{k ^{2} - 1}

n = 1 \sum \infty \frac{1 + cos ( n )}{n}

n = 0 \sum \infty 1 - e^{- n}

What is the sum from n=1 to infinity of 6/(3n+2) ?
The sum from n=1 to infinity of 6/(3n+2) is diverges