What is the sum from n=1 to infinity of (4n)/(3n+1) ?

The sum from n=1 to infinity of (4n)/(3n+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 (4n)/(3n+1)

n = 1 \sum \infty \frac{4 n}{3 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{4 n}{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{n}{3 n + 1}

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

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

n = 1 \sum \infty \frac{1 1 ^{n}}{n !}

n = 0 \sum \infty \frac{i}{n ^{2}}

k = 1 \sum \infty (cos (14))^{k}

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

What is the sum from n=1 to infinity of (4n)/(3n+1) ?
The sum from n=1 to infinity of (4n)/(3n+1) is diverges