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

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

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

n = 1 \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

k = 1 \sum \infty \frac{k ^{2}}{2 ^{k}}

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

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

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

n = 1 \sum \infty tan (\frac{3}{k ^{2}})

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