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

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

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-n)

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

n = 1 \sum \infty \frac{1}{8 n + 4}

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

n = 0 \sum \infty \frac{n}{ln ( n )}

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

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

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