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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{4 n ^{2}}

n = 0 \sum \infty 8 \cdot (\frac{2}{7})^{n}

k = 1 \sum \infty k e^{- 7 k^{2}}

n = 0 \sum \infty \frac{n !}{n}

n = 0 \sum \infty \frac{2 n}{n ^{3}}

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