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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 2 \sum \infty (- 0.25)^{n}

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

n = 1 \sum \infty sin (\frac{nπ}{2})

n = 0 \sum \infty (\frac{4}{5})^{2 n}

n = 1 \sum \infty 5 ln (\frac{n}{n + 1})

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