What is the sum from n=0 to infinity of (-3)/(n^2-4) ?

The sum from n=0 to infinity of (-3)/(n^2-4) is converges

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sum from n=0 to infinity of (-3)/(n^2-4)

n = 0 \sum \infty \frac{- 3}{n ^{2} - 4}

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= - 3 converges

= converges

n = 0 \sum \infty \frac{1}{n ^{100}}

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

m = 2 \sum \infty \frac{2}{5 ^{m}}

n = 1 \sum \infty \frac{1}{n ^{\frac{10}{8}}}

n = 1 \sum \infty \frac{7 n}{n !}

What is the sum from n=0 to infinity of (-3)/(n^2-4) ?
The sum from n=0 to infinity of (-3)/(n^2-4) is converges