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

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

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

n = 3 \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

k = 1 \sum \infty \frac{1}{k + 3}

n = 1 \sum \infty \frac{1}{2 ^{2 n + 1}}

n = 2 \sum \infty \frac{6}{3 ^{2 n}}

n = 0 \sum \infty \frac{ln ( n ^{2} )}{n}

n = 0 \sum \infty \frac{3}{3 + n ^{2}}

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