What is the sum from n=1 to infinity of (2n)/(7n-4) ?

The sum from n=1 to infinity of (2n)/(7n-4) 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 (2n)/(7n-4)

n = 1 \sum \infty \frac{2 n}{7 n - 4}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{7 n - 4}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{7 n - 4}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 2 \sum \infty \frac{5}{n ^{2} - n}

n = 1 \sum \infty \frac{n ^{2}}{n ^{4}}

n = 5 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

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

n = 0 \sum \infty \frac{5}{6 ^{n}}

What is the sum from n=1 to infinity of (2n)/(7n-4) ?
The sum from n=1 to infinity of (2n)/(7n-4) is diverges