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}

\frac{d}{d x} (\frac{3}{4} x^{2})

\int (9 - y^{2})^{2} d y

x \to 0 lim (\frac{3 x ^{5} - 3 x ^{3} + 4 x}{x ^{3} + 6 x ^{2} - 2 x})

\int \frac{- 2 cos ( x )}{sin ^{2} ( x )} d x

\frac{d}{d x} (x e^{\frac{1}{2}})

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