What is the sum from n=6 to infinity of 1/(n^2-4) ?

The sum from n=6 to infinity of 1/(n^2-4) is converges

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sum from n=6 to infinity of 1/(n^2-4)

n = 6 \sum \infty \frac{1}{n ^{2} - 4}

converges

Solution steps

n = 6 \sum \infty \frac{1}{n ^{2} - 4}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ^{- 10}}

y = x^{2} + x + 3

\frac{1}{x}, at x = 9

\frac{d}{d x} (x (x - 1))

x \to 0 + lim (tan^{x} (5 x))

What is the sum from n=6 to infinity of 1/(n^2-4) ?
The sum from n=6 to infinity of 1/(n^2-4) is converges