What is the sum from n=1 to infinity of (6n)/(3n+9) ?

The sum from n=1 to infinity of (6n)/(3n+9) 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 (6n)/(3n+9)

n = 1 \sum \infty \frac{6 n}{3 n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{6 n}{3 n + 9}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{n}{3 n + 9}

Apply Series Divergence Test:

diverges

= 6 diverges

= diverges

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

n = 0 \sum \infty cos (\frac{nπ}{3})

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

n = 1 \sum \infty - 8 (\frac{4}{5})^{n - 1}

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

What is the sum from n=1 to infinity of (6n)/(3n+9) ?
The sum from n=1 to infinity of (6n)/(3n+9) is diverges