What is the sum from n=0 to infinity of 1/(16n^2-4) ?

The sum from n=0 to infinity of 1/(16n^2-4) is converges

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sum from n=0 to infinity of 1/(16n^2-4)

n = 0 \sum \infty \frac{1}{16 n ^{2} - 4}

converges

Solution steps

n = 0 \sum \infty \frac{1}{16 n ^{2} - 4}

Apply Series Integral Test:

converges

= converges

n = 1 \sum \infty (0.7)^{n - 1}

n = 0 \sum \infty \frac{1}{n + 2} - \frac{1}{n}

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

n = 1 \sum \infty \frac{1}{5 n + 3}

n = - \infty \sum \infty n

What is the sum from n=0 to infinity of 1/(16n^2-4) ?
The sum from n=0 to infinity of 1/(16n^2-4) is converges