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

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

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

n = 2 \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{n}{9 ^{n}}

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

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

i = 1 \sum \infty \frac{1}{3 ^{i}}

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