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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 4 converges

= converges

n = 0 \sum \infty \frac{i ^{2 n}}{n}

n = 2 \sum \infty (- 0.9)^{n}

n = 1 \sum \infty \frac{n !}{2 n}

n = 1 \sum \infty n^{- 0.6}

n = 0 \sum \infty \frac{4}{9 - n ^{2}}

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