What is the sum from n=3 to infinity of 4/(n^2-n) ?

The sum from n=3 to infinity of 4/(n^2-n) is 2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=3 to infinity of 4/(n^2-n)

n = 3 \sum \infty \frac{4}{n ^{2} - n}

2

Solution steps

n = 3 \sum \infty \frac{4}{n ^{2} - n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 3 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

\frac{1}{2}

= 4 \cdot \frac{1}{2}

Simplify

4 \cdot \frac{1}{2} : 2

= 2

n = 0 \sum \infty \frac{1}{n ^{2} + 5 n + 4}

n = 3 \sum \infty (\frac{2}{3})^{n}

n = 1 \sum \infty \frac{1}{n ^{\frac{4}{5}}}

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

m = 2 \sum \infty \frac{6}{7 ^{m}}

What is the sum from n=3 to infinity of 4/(n^2-n) ?
The sum from n=3 to infinity of 4/(n^2-n) is 2