What is the sum from n=1 to infinity of (6n)/(8n+9) ?

The sum from n=1 to infinity of (6n)/(8n+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)/(8n+9)

n = 1 \sum \infty \frac{6 n}{8 n + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{6 n}{8 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}{8 n + 9}

Apply Series Divergence Test:

diverges

= 6 diverges

= diverges

\int p^{8} ln (p) d p

\int cos^{3} (8 x) sin^{- 2} (8 x) d x

y^{^{'}} = \frac{( y ^{2} + 1 )}{2 y}

\int \frac{12 x ^{2} + x + 2}{x ( x ^{2} + 1 )} d x

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

What is the sum from n=1 to infinity of (6n)/(8n+9) ?
The sum from n=1 to infinity of (6n)/(8n+9) is diverges