What is the sum from n=1 to infinity of (6n)/(4n+1) ?

The sum from n=1 to infinity of (6n)/(4n+1) 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)/(4n+1)

n = 1 \sum \infty \frac{6 n}{4 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{6 n}{4 n + 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{n}{4 n + 1}

Apply Series Divergence Test:

diverges

= 6 diverges

= diverges

n = 1 \sum \infty \frac{2}{5 n ^{2} + 5 n}

n = 0 \sum \infty (\frac{5}{12})^{n}

n = 0 \sum \infty \frac{3}{5 ^{n}} + \frac{2}{n}

n = 0 \sum \infty \frac{( - 1 )}{n}

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

What is the sum from n=1 to infinity of (6n)/(4n+1) ?
The sum from n=1 to infinity of (6n)/(4n+1) is diverges