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

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

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

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

- \frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

- \frac{1}{2}

= - \frac{1}{2}

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