What is the sum from n=1 to infinity of 1/(4n^2-81) ?

The sum from n=1 to infinity of 1/(4n^2-81) is converges

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sum from n=1 to infinity of 1/(4n^2-81)

n = 1 \sum \infty \frac{1}{4 n ^{2} - 81}

converges

Solution steps

n = 1 \sum \infty \frac{1}{4 n ^{2} - 81}

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{6}}}

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

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

n = 0 \sum \infty \frac{1}{7 n + 1}

n = 1 \sum \infty \frac{2}{3}

What is the sum from n=1 to infinity of 1/(4n^2-81) ?
The sum from n=1 to infinity of 1/(4n^2-81) is converges