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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 3 converges

= converges

n = 1 \sum \infty (\frac{5}{8})^{n - 1}

k = 2 \sum \infty (- 0.4)^{k}

n = 0 \sum \infty 7 e^{- 7 n}

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

k = 4 \sum \infty \frac{3}{k ln ^{2} ( k )}

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