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

The sum from n=2 to infinity of 1/(4n^2-1) is 1/6

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

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

\frac{1}{6}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{6}

= \frac{1}{6}

\frac{\partial}{\partial y} (y e^{3 x})

n = 0 \sum \infty x^{n + 1}

\int \frac{1}{x 2 x + 1} d x

\frac{d}{d x} (\frac{v}{x})

What is the sum from n=2 to infinity of 1/(4n^2-1) ?
The sum from n=2 to infinity of 1/(4n^2-1) is 1/6