What is the sum from n=1 to infinity of 3/((4n^2-1)) ?

The sum from n=1 to infinity of 3/((4n^2-1)) is 3/2

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sum from n=1 to infinity of 3/((4n^2-1))

n = 1 \sum \infty \frac{3}{( 4 n ^{2} - 1 )}

\frac{3}{2}

Solution steps

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 3 \cdot \frac{1}{2}

Simplify

3 \cdot \frac{1}{2} : \frac{3}{2}

= \frac{3}{2}

\frac{\partial}{\partial y} ((2 x + 3 y)^{10})

x y^{^{'}} = y - x^{2} + y^{2}

x y^{^{'}} = 2 y

n = 0 \sum \infty - 5

\int \frac{4 x ^{3} + 3}{x ^{4} + 3 x} d x

What is the sum from n=1 to infinity of 3/((4n^2-1)) ?
The sum from n=1 to infinity of 3/((4n^2-1)) is 3/2