What is the sum from n=0 to infinity of (2n)/(5n+1) ?

The sum from n=0 to infinity of (2n)/(5n+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=0 to infinity of (2n)/(5n+1)

n = 0 \sum \infty \frac{2 n}{5 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{2 n}{5 n + 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{n}{5 n + 1}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty n e^{- \frac{n}{9}}

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

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

k = 9 \sum \infty \frac{7}{k ^{6.6}}

n = 1 \sum \infty \frac{n}{n + 11}

What is the sum from n=0 to infinity of (2n)/(5n+1) ?
The sum from n=0 to infinity of (2n)/(5n+1) is diverges