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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{2}{n ^{0.85}}

n = 1 \sum \infty (- \frac{6}{7})^{n - 1}

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

n = 1 \sum \infty \frac{( e ^{\frac{1}{n}} )}{n}

n = 1 \sum \infty \frac{1}{n ln ( n ^{2} )}

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