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

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

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 1/(n^3+n)

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{3} + 8}

n = 1 \sum \infty \frac{n}{n + 10}

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

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

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

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