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

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

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

\int_{0}^{\frac{π}{2}} cos (t) d t

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 = 0

\frac{d ^{2} x}{d t ^{2}} + 4 x = sin (1 t), x (0) = 1

\frac{d}{d x} (- \frac{1}{15} cos (15 x))

\int e^{2 - x} d x

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