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

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

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

x \to 0 + lim ((x)^{x})

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\frac{\partial}{\partial x} (\frac{2 y}{( x ^{2} + y ^{2} )})

x \to 0 + lim (cos (x) - 1)

\int 2 sec (x) tan (x) d x

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