What is the sum from n=0 to infinity of 3/(n^2+3n+2) ?

The sum from n=0 to infinity of 3/(n^2+3n+2) is 3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 3/(n^2+3n+2)

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

3

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

Apply Telescoping Series Test:

1

= 3 \cdot 1

Simplify

= 3

n = 1 \sum \infty 4 (\frac{1}{3})^{n - 1}

n = 1 \sum \infty 3 n^{2} e^{- n^{3}}

n = 1 \sum \infty [53 (\frac{1}{2})^{n}]

n = 0 \sum \infty \frac{8}{n ( n + 2 )}

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

What is the sum from n=0 to infinity of 3/(n^2+3n+2) ?
The sum from n=0 to infinity of 3/(n^2+3n+2) is 3