What is the sum from n=1 to infinity of (4n)/(7n+1) ?

The sum from n=1 to infinity of (4n)/(7n+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)/(7n+1)

n = 1 \sum \infty \frac{4 n}{7 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{4 n}{7 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}{7 n + 1}

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

n = 1 \sum \infty 2^{n - 1}

\int_{0}^{\frac{π}{6}} (1 - cos (3 x)) sin (3 x) d x

4 e^{2 x}

\frac{d}{d x} (3 x^{2} + 1)

n = 0 \sum \infty \frac{1}{4}

What is the sum from n=1 to infinity of (4n)/(7n+1) ?
The sum from n=1 to infinity of (4n)/(7n+1) is diverges