What is the sum from n=3 to infinity of 1/(n^2-4) ?

The sum from n=3 to infinity of 1/(n^2-4) is converges

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sum from n=3 to infinity of 1/(n^2-4)

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

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

n = 1 \sum \infty (\frac{2}{13})^{n}

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

n = 0 \sum \infty (4 + n^{2})^{- 1}

k = 0 \sum \infty 1^{k}

What is the sum from n=3 to infinity of 1/(n^2-4) ?
The sum from n=3 to infinity of 1/(n^2-4) is converges