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

n = 0 \sum \infty \frac{1}{n ^{\frac{5}{2}}}

n = 0 \sum \infty \frac{n}{e ^{7 n}}

n = 4 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}

n = 2 \sum \infty \frac{5}{n ^{3}}

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